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How to Solve a Quadratic Equation: A Step-by-Step Guide

Last Updated: May 27, 2024 Fact Checked

Factoring the Equation

Using the quadratic formula, completing the square, practice problems and answers, expert q&a.

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 9 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,421,140 times.

A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. [1] X Research source There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.

Quadradic Formula for Solving Equations

• Then, use the process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4. [3] X Research source

• 3x = -1 ..... by subtracting
• 3x/3 = -1/3 ..... by dividing
• x = -1/3 ..... simplified
• x = 4 ..... by subtracting
• x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.

• So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.

• 4x 2 - 5x - 13 = x 2 -5
• 4x 2 - x 2 - 5x - 13 +5 = 0
• 3x 2 - 5x - 8 = 0

• {-b +/-√ (b 2 - 4ac)}/2
• {-(-5) +/-√ ((-5) 2 - 4(3)(-8))}/2(3) =
• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3)

• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) =
• {5 +/-√(25 + 96)}/6
• {5 +/-√(121)}/6

• (5 + 11)/6 = 16/6
• (5-11)/6 = -6/6

• x = (-1, 8/3)

• 2x 2 - 9 = 12x =
• In this equation, the a term is 2, the b term is -12, and the c term is -9.

• 2x 2 - 12x - 9 = 0
• 2x 2 - 12x = 9

• 2x 2 /2 - 12x/2 = 9/2 =
• x 2 - 6x = 9/2

• -6/2 = -3 =
• (-3) 2 = 9 =
• x 2 - 6x + 9 = 9/2 + 9

• x = 3 + 3(√6)/2
• x = 3 - 3(√6)/2)

• If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example: [14] X Research source Thanks Helpful 0 Not Helpful 0
• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Thanks Helpful 3 Not Helpful 0
• As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. Thanks Helpful 1 Not Helpful 0

You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
• ↑ https://www.mathsisfun.com/algebra/factoring-quadratics.html
• ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
• ↑ https://www.cuemath.com/algebra/quadratic-equations/
• ↑ https://www.purplemath.com/modules/solvquad4.htm
• ↑ http://www.purplemath.com/modules/quadform.htm
• ↑ https://uniskills.library.curtin.edu.au/numeracy/algebra/quadratic-equations/
• ↑ http://www.mathsisfun.com/algebra/completing-square.html
• ↑ https://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm

About This Article

To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down! Did this summary help you? Yes No

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Quadratic Equation

Quadratic equations are second-degree algebraic expressions and are of the form ax 2 + bx + c = 0. The term "quadratic" comes from the Latin word "quadratus" meaning square, which refers to the fact that the variable x is squared in the equation. In other words, a quadratic equation is an “equation of degree 2.” There are many scenarios where a quadratic equation is used. Did you know that when a rocket is launched, its path is described by a quadratic equation? Further, a quadratic equation has numerous applications in physics, engineering, astronomy, etc.

Quadratic equations have maximum of two solutions, which can be real or complex numbers. These two solutions (values of x) are also called the roots of the quadratic equations and are designated as (α, β). We shall learn more about the roots of a quadratic equation in the below content.

What is Quadratic Equation?

A quadratic equation is an algebraic equation of the second degree in x. The quadratic equation in its standard form is ax 2 + bx + c = 0, where a and b are the coefficients, x is the variable, and c is the constant term. The important condition for an equation to be a quadratic equation is the coefficient of x 2 is a non-zero term (a ≠ 0). For writing a quadratic equation in standard form, the x 2 term is written first, followed by the x term, and finally, the constant term is written.

Further, in real math problems the quadratic equations are presented in different forms: (x - 1)(x + 2) = 0, -x 2 = -3x + 1, 5x(x + 3) = 12x, x 3 = x(x 2 + x - 3). All of these equations need to be transformed into standard form of the quadratic equation before performing further operations.

Roots of a Quadratic Equation

The roots of a quadratic equation are the two values of x, which are obtained by solving the quadratic equation. These roots of the quadratic equation are also called the zeros of the equation. For example, the roots of the equation x 2 - 3x - 4 = 0 are x = -1 and x = 4 because each of them satisfies the equation. i.e.,

• At x = -1, (-1) 2 - 3(-1) - 4 = 1 + 3 - 4 = 0
• At x = 4, (4) 2 - 3(4) - 4 = 16 - 12 - 4 = 0

There are various methods to find the roots of a quadratic equation. The usage of the quadratic formula is one of them.

Quadratic Formula

Quadratic formula is the simplest way to find the roots of a quadratic equation . There are certain quadratic equations that cannot be easily factorized, and here we can conveniently use this quadratic formula to find the roots in the quickest possible way. The two roots in the quadratic formula are presented as a single expression. The positive sign and the negative sign can be alternatively used to obtain the two distinct roots of the equation.

Quadratic Formula: The roots of a quadratic equation ax 2 + bx + c = 0 are given by x = [-b ± √(b 2 - 4ac)]/2a.

This formula is also known as the Sridharacharya formula .

Example: Let us find the roots of the same equation that was mentioned in the earlier section x 2 - 3x - 4 = 0 using the quadratic formula.

a = 1, b = -3, and c = -4.

x = [-b ± √(b 2 - 4ac)]/2a = [-(-3) ± √((-3) 2 - 4(1)(-4))]/2(1) = [3 ± √25] / 2 = [3 ± 5] / 2 = (3 + 5)/2 or (3 - 5)/2 = 8/2 or -2/2 = 4 or -1 are the roots.

Proof of Quadratic Formula

Consider an arbitrary quadratic equation: ax 2 + bx + c = 0, a ≠ 0

To determine the roots of this equation, we proceed as follows:

ax 2 + bx = -c ⇒ x 2 + bx/a = -c/a

Now, we express the left-hand side as a perfect square , by introducing a new term (b/2a) 2 on both sides:

• x 2 + bx/a + (b/2a) 2 = -c/a + (b/2a) 2

The left-hand side is now a perfect square:

(x + b/2a) 2 = -c/a + b 2 /4a 2 ⇒ (x + b/2a) 2 = (b 2 - 4ac)/4a 2

This is good for us, because now we can take square roots to obtain:

x + b/2a = ±√(b 2 - 4ac)/2a

x = (-b ± √(b 2 - 4ac))/2a

Thus, by completing the squares, we were able to isolate x and obtain the two roots of the equation.

Nature of Roots of the Quadratic Equation

The roots of a quadratic equation are usually represented to by the symbols alpha (α), and beta (β). Here we shall learn more about how to find the nature of roots of a quadratic equation without actually finding the roots of the equation.

The nature of roots of a quadratic equation can be found without actually finding the roots (α, β) of the equation. This is possible by taking the discriminant value, which is part of the formula to solve the quadratic equation. The value b 2 - 4ac is called the discriminant of a quadratic equation and is designated as 'D'. Based on the discriminant value the nature of the roots of the quadratic equation can be predicted.

Discriminant: D = b 2 - 4ac

• D > 0, the roots are real and distinct
• D = 0, the roots are real and equal.
• D < 0, the roots do not exist or the roots are imaginary .

Now, check out the formulas to find the sum and the product of the roots of the equation.

Sum and Product of Roots of Quadratic Equation

The coefficient of x 2 , x term, and the constant term of the quadratic equation ax 2 + bx + c = 0 are useful in determining the sum and product of the roots of the quadratic equation. The sum and product of the roots of a quadratic equation can be directly calculated from the equation, without actually finding the roots of the quadratic equation. For a quadratic equation ax 2 + bx + c = 0, the sum and product of the roots are as follows.

• Sum of the Roots: α + β = -b/a = - Coefficient of x/ Coefficient of x 2
• Product of the Roots: αβ = c/a = Constant term/ Coefficient of x 2

Writing Quadratic Equations Using Roots

The quadratic equation can also be formed for the given roots of the equation. If α, β, are the roots of the quadratic equation, then the quadratic equation is as follows.

x 2 - (α + β)x + αβ = 0

Example: What is the quadratic equation whose roots are 4 and -1?

Solution: It is given that α = 4 and β = -1. The corresponding quadratic equation is found by:

x 2 - (α + β)x + αβ = 0 x 2 - (α + β)x + αβ = 0 x 2 - (4 - 1)x + (4)(-1) = 0 x 2 - 3x - 4 = 0

Formulas Related to Quadratic Equations

The following list of important formulas is helpful to solve quadratic equations.

• The quadratic equation in its standard form is ax 2 + bx + c = 0
• For D > 0 the roots are real and distinct.
• For D = 0 the roots are real and equal.
• For D < 0 the real roots do not exist, or the roots are imaginary.
• The formula to find the roots of the quadratic equation is x = [-b ± √(b 2 - 4ac)]/2a.
• The sum of the roots of a quadratic equation is α + β = -b/a.
• The product of the Root of the quadratic equation is αβ = c/a.
• The quadratic equation whose roots are α, β, is x 2 - (α + β)x + αβ = 0.
• The condition for the quadratic equations a 1 x 2 + b 1 x + c 1 = 0, and a 2 x 2 + b 2 x + c 2 = 0 having the same roots is (a 1 b 2 - a 2 b 1 ) (b 1 c 2 - b 2 c 1 ) = (a 2 c 1 - a 1 c 2 ) 2 .
• When a > 0, the quadratic expression f(x) = ax 2 + bx + c has a minimum value at x = -b/2a.
• When a < 0, the quadratic expression f(x) = ax 2 + bx + c has a maximum value at x = -b/2a.
• The domain of any quadratic function is the set of all real numbers.

Methods to Solve Quadratic Equations

A quadratic equation can be solved to obtain two values of x or the two roots of the equation. There are four different methods to find the roots of the quadratic equation. The four methods of solving the quadratic equations are as follows.

• Factorizing of Quadratic Equation
• Using quadratic formula (which we have seen already)

Method of Completing the Square

• Graphing Method to Find the Roots

Let us look in detail at each of the above methods to understand how to use these methods, their applications, and their uses.

Solving Quadratic Equations by Factorization

Factorization of quadratic equation follows a sequence of steps. For a general form of the quadratic equation ax 2 + bx + c = 0, we need to first split the middle term into two terms, such that the product of the terms is equal to the constant term. Further, we can take the common terms from the available term, to finally obtain the required factors as follows:

• x 2 + (a + b)x + ab = 0
• x 2 + ax + bx + ab = 0
• x(x + a) + b(x + a)
• (x + a)(x + b) = 0

Here is an example to understand the factorization process.

• x 2 + 5x + 6 = 0
• x 2 + 2x + 3x + 6 = 0
• x(x + 2) + 3(x + 2) = 0
• (x + 2)(x + 3) = 0

Thus the two obtained factors of the quadratic equation are (x + 2) and (x + 3). To find its roots, just set each factor to zero and solve for x. i.e., x + 2 = 0 and x + 3 = 0 which gives x = -2 and x = -3. Thus, x = -2 and x = -3 are the roots of x 2 + 5x + 6 = 0.

Further, there is another important method of solving a quadratic equation. The method of completing the square for a quadratic equation is also useful to find the roots of the equation.

The method of completing the square in a quadratic equation is to algebraically square and simplify, to obtain the required roots of the equation. Consider a quadratic equation ax 2 + bx + c = 0, a ≠ 0. To determine the roots of this equation, we simplify it as follows:

• ax 2 + bx + c = 0
• ax 2 + bx = -c
• x 2 + bx/a = -c/a

Now, we express the left-hand side as a perfect square, by introducing a new term (b/2a) 2 on both sides:

• (x + b/2a) 2 = -c/a + b 2 /4a 2
• (x + b/2a) 2 = (b 2 - 4ac)/4a 2
• x + b/2a = + √(b 2 - 4ac)/2a
• x = - b/2a + √(b 2 - 4ac)/2a
• x = [-b ± √(b 2 - 4ac)]/2a

Here the '+' sign gives one root and the '-' sign gives another root of the quadratic equation. Generally, this detailed method is avoided, and only the quadratic formula is used to obtain the required roots.

Graphing a Quadratic Equation

The point(s) where the graph cuts the horizontal x-axis (typically the x-intercepts ) is the solution of the quadratic equation. These points can also be algebraically obtained by equalizing the y value to 0 in the function y = ax 2 + bx + c and solving for x.

Quadratic Equations Having Common Roots

Consider two quadratic equations having common roots a 1 x 2 + b 1 x + c 1 = 0, and a 2 x 2 + b 2 x + c 2 = 0. Let us solve these two equations to find the conditions for which these equations have a common root. The two equations are solved for x 2 and x respectively.

(x 2 )(b 1 c 2 - b 2 c 1 ) = (-x)/(a 1 c 2 - a 2 c 1 ) = 1/(a 1 b 2 - a 2 b 1 )

x 2 = (b 1 c 2 - b 2 c 1 ) / (a 1 b 2 - a 2 b 1 )

x = (a 2 c 1 - a 1 c 2 ) / (a 1 b 2 - a 2 b 1 )

Hence, by simplifying the above two expressions we have the following condition for the two equations having the common root.

(a 1 b 2 - a 2 b 1 ) (b 1 c 2 - b 2 c 1 ) = (a 2 c 1 - a 1 c 2 ) 2

Maximum and Minimum Value of Quadratic Expression

The maximum and minimum values of the quadratic expressions are of further help to find the range of the quadratic expression: The range of the quadratic expressions also depends on the value of a. For positive values of a( a > 0), the range is [ F(-b/2a), ∞), and for negative values of a ( a < 0), the range is (-∞, F(-b/2a)].

• For a > 0, Range: [ f(-b/2a), ∞)
• For a < 0, Range: (-∞, f(-b/2a)]

Note that the domain of a quadratic function is the set of all real numbers, i.e., (-∞, ∞).

Tips and Tricks on Quadratic Equation:

Some of the below-given tips and tricks on quadratic equations are helpful to more easily solve quadratic equations.

• The quadratic equations are generally solved through factorization. But in instances when it cannot be solved by factorization, the quadratic formula is used.
• The roots of a quadratic equation are also called the zeroes of the equation.
• For quadratic equations having negative discriminant values, the roots are represented by complex numbers.
• The sum and product of the roots of a quadratic equation can be used to find higher algebraic expressions involving these roots.

☛Related Topics:

• Roots Calculator
• Quadratic Factoring Calculator
• Roots of Quadratic Equation Calculator

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Quadratic Equations Examples

Example 1: James is a fitness enthusiast and goes for a jog every morning. The park where he jogs is rectangular in shape and measures 12m × 8m. An environmentalist group plans to revamp the park and decides to build a pathway surrounding the park. This would increase the total area to 140 sq m. What will be the width of the pathway?

Let’s denote the width of the pathway by x.

Then, the length and breadth of the outer rectangle is (12+2x) m and (8+2x) m.

Given that, area = 140

(12 + 2x)(8 + 2x) = 140

2(6 + x) 2(4 + x) = 140

(6 + x)(4 + x) = 35

24 + 6x + 4x + x 2 = 35

x 2 + 10x -11 = 0

x 2 + 11x - x - 11 = 0

x(x + 11) - 1(x + 11) = 0

(x + 11)(x - 1) = 0

(x + 11) =0 and (x - 1) = 0

x = -11 and x = 1

Since length can’t be negative, we take x = 1.

Answer: Therefore the width of the pathway is 1 m.

Example 2: Rita throws a ball upwards from a platform that is 20m above the ground. The height of the ball from the ground at a time 't', is denoted by 'h'. Suppose h = -4t 2 + 16t + 20. Find the maximum height attained by the ball.

We can rearrange the terms of the quadratic equation

h = -4t 2 + 16t + 20

in such a way that it is easy to find the maximum value of this equation.

= -4t 2 + 16t + 20

= -4(t 2 - 4t - 5)

= -4((t - 2) 2 - 9)

= -4(t - 2) 2 + 36

We should keep the value of (t - 2) 2 minimum in order to find the maximum value of h.

So, the minimum value (t - 2) 2 can take is 0.

Answer: Therefore the maximum height attained is 36m.

Example 3: Find the quadratic equation having the roots 5 and 8 respectively.

The quadratic equation having the roots α, β, is x 2 - (α + β)x + αβ = 0.

Given α = 5, and β = 8.

Therefore the quadratic equation is:

x 2 - (5 + 8)x + 5×8 = 0

x 2 - 13x + 40 = 0

Answer: Hence the required quadratic equation is x 2 - 13x + 40 = 0

Example 4: The quad equation 2x 2 + 9x + 7 = 0 has roots α, β. Find the quadratic equation having the roots 1/α, and 1/β.

The quadratic equation having roots that are reciprocal to the roots of the equation ax 2 + bx + c = 0, is cx 2 + bx + a = 0.

The given quadratic equation is 2x 2 + 9x + 7 = 0.

Hence the required equation having reciprocal roots is 7x 2 + 9x + 2 = 0.

Method 2: From the given equation,

α + β = -9/2 and α β = 2/7.

The new equation should have its roots to be 1/α and 1/β.

Their sum = 1/α + 1/β = (α + β) / α β = -9/7 Their product = 1/α β = 2/7 Thus, the required equation is, x 2 - (1/α + 1/β)x + 1/α β = 0 x 2 - (-9/7)x + 2/7 = 0 Multiplying both sides by 7, 7x 2 + 9x + 2 = 0

Answer: Therefore the equation is 7x 2 + 9x + 2 = 0.

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Practice Questions on Quadratic Equation

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FAQs on Quadratic Equation

What is the definition of a quadratic equation.

A quadratic equation in math is a second-degree equation of the form ax 2 + bx + c = 0. Here a and b are the coefficients, c is the constant term, and x is the variable. Since the variable x is of the second degree, there are two roots or answers for this quadratic equation. The roots of the quadratic equation can be found by either solving by factorizing or through the use of the quadratic formula.

What is the Quadratic Formula?

The quadratic equation formula to solve the equation ax 2 + bx + c = 0 is x = [-b ± √(b 2 - 4ac)]/2a. Here we obtain the two values of x, by applying the plus and minus symbols in this formula. Hence the two possible values of x are [-b + √(b 2 - 4ac)]/2a, and [-b - √(b 2 - 4ac)]/2a.

How do You Solve a Quadratic Equation?

There are several methods to solve quadratic equations, but the most common ones are factoring, using the quadratic formula, and completing the square.

• Factoring involves finding two numbers that multiply to equal the constant term, c, and add up to the coefficient of x, b.
• The quadratic formula is used when factoring is not possible, and it is given by x = [-b ± √(b 2 - 4ac)]/2a.
• Completing the square involves rewriting the quadratic equation in a different form that allows you to easily solve for x.

What is Determinant in Quadratic Formula?

The value b 2 - 4ac is called the discriminant and is designated as D. The discriminant is part of the quadratic formula. The discriminants help us to find the nature of the roots of the quadratic equation, without actually finding the roots of the quadratic equation.

What are Some Real-Life Applications of Quadratic Equations?

Quadratic equations are used to find the zeroes of the parabola and its axis of symmetry . There are many real-world applications of quadratic equations.

• They can be used in running time problems to evaluate the speed, distance or time while traveling by car, train or plane.
• Quadratic equations describe the relationship between quantity and the price of a commodity.
• Similarly, demand and cost calculations are also considered quadratic equation problems.
• It can also be noted that a satellite dish or a reflecting telescope has a shape that is defined by a quadratic equation.

How are Quadratic Equations Different From Linear Equations?

A linear degree is an equation of a single degree and one variable, and a quadratic equation is an equation in two degrees and a single variable. A linear equation is of the the form ax + b = 0 and a quadratic equation is of the form ax 2 + bx + c = 0. A linear equation has a single root and a quadratic equation has two roots or two answers. Also, a quadratic equation is a product of two linear equations.

What Are the 4 Ways To Solve A Quadratic Equation?

The four ways of solving a quadratic equation are as follows.

• Factorizing method
• Roots of Quadratic Equation Formula Method
• Method of Completing Squares
• Graphing Method

How to Solve a Quadratic Equation by Completing the Square?

The quadratic equation is solved by the method of completing the square and it uses the formula (a + b)^2 = a 2 + 2ab + b 2 (or) (a - b)^2 = a 2 - 2ab + b 2 .

How to Find the Value of the Discriminant?

The value of the discriminant in a quadratic equation can be found from the variables and constant terms of the standard form of the quadratic equation ax 2 + bx + c = 0. The value of the discriminant is D = b 2 - 4ac, and it helps to predict the nature of roots of the quadratic equation, without actually finding the roots of the equation.

How Do You Solve Quadratic Equations With Graphing?

The quadratic equation can be solved similarly to a linear equal by graphing. Let us take the quadratic equation ax 2 + bx + c = 0 as y = ax 2 + bx + c . Here we take the set of values of x and y and plot the graph. The two points where this graph meets the x-axis, are the solutions of this quadratic equation.

How Important Is the Discriminant of a Quadratic Equation?

The discriminant is very much needed to easily find the nature of the roots of the quadratic equation. Without the discriminant, finding the nature of the roots of the equation is a long process, as we first need to solve the equation to find both the roots. Hence the discriminant is an important and needed quantity, which helps to easily find the nature of the roots of the quadratic equation.

Where Can I Find Quadratic Equation Solver?

To get the quadratic equation solver, click here . Here, we can enter the values of a, b, and c for the quadratic equation ax 2 + bx + c = 0, then it will give you the roots along with a step-by-step procedure.

What is the Use of Discriminants in Quadratic Formula?

The discriminant (D = b 2 - 4ac) is useful to predict the nature of the roots of the quadratic equation. For D > 0, the roots are real and distinct, for D = 0 the roots are real and equal, and for D < 0, the roots do not exist or the roots are imaginary complex numbers . With the help of this discriminant and with the least calculations, we can find the nature of the roots of the quadratic equation.

How do you Solve a Quadratic Equation without Using the Quadratic Formula?

There are two alternative methods to the quadratic formula. One method is to solve the quadratic equation through factorization, and another method is by completing the squares. In total there are three methods to find the roots of a quadratic equation.

How to Derive Quadratic Formula?

The algebra formula (a + b) 2 = a 2 + 2ab + b 2 is used to solve the quadratic equation and derive the quadratic formula. This algebraic formula is used to manipulate the quadratic equation and derive the quadratic formula to find the roots of the equation.

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The most commonly used methods for solving quadratic equations are:

1 . Factoring method

2 . Solving quadratic equations by completing the square

3 . Using quadratic formula

In the following sections, we'll go over these methods.

Method 1A : Factoring method

If a quadratic trinomial can be factored, this is the best solving method.

We often use this method when the leading coefficient is equal to 1 or -1. If this is not the case, then it is better to use some other method.

Example 01: Solve $ x^2 \color{red}{-8}x \color{blue}{+ 15} = 0 $ by factoring.

Here we see that the leading coefficient is 1, so the factoring method is our first choice.

To factor this equation, we must find two numbers ( $ a $ and $ b $ ) with a sum is $ a + b = \color{red}{8} $ and a product of $ a \cdot b = \color{blue}{15} $.

After some trials and errors, we see that $ a = 3 $ and $ b = 5 $.

Now we use formula $ x^2 - 8x + 15 = (x - a)(x - b) $ to get factored form:

Divide the factored form into two linear equations to get solutions.

Method 1B : Factoring - special cases

Example 02: Solve $ x^2 -8x = 0 $ by factoring.

In this case, (when the coefficient c = 0 ) we can factor out $ \color{blue}{x} $ out of $ x^2 - 8x $.

Example 03: Solve $ x^2 - 16 = 0 $ by factoring.

In this case, ( when the middle term is equal 0) we can use the difference of squares formula.

Method 3 : Solve using quadratic formula

This method solves all types of quadratic equations. It works best when solutions contain some radicals or complex numbers.

Example 05: Solve equation $ 2x^2 + 3x - 2 = 0$ by using quadratic formula.

Step 1 : Read the values of $a$, $b$, and $c$ from the quadratic equation. ( $a$ is the number in front of $x^2$ , $b$ is the number in front of $x$ and $c$ is the number at the end)

Step 2 :Plug the values for a, b, and c into the quadratic formula and simplify.

Step 3 : Solve for $x_1$ and $x_2$

Method 2 : Completing the square

This method can be used to solve all types of quadratic equations, although it can be complicated for some types of equations. The method involves seven steps.

Example 04: Solve equation $ 2x^2 + 8x - 10= 0$ by completing the square.

Step 1 : Divide the equation by the number in front of the square term.

Step 2 : move $-5$ to the right:

Step 3 : Take half of the x-term coefficient $ \color{blue}{\dfrac{4}{2}} $, square it $ \color{blue}{\left(\dfrac{4}{2} \right)^2} $ and add this value to both sides.

Step 4 : Simplify left and right side.

Step 5 : Write the perfect square on the left.

Step 6 : Take the square root of both sides.

Step 7 : Solve for $x_1$ and $x_2$ .

1. Quadratic Equation — step-by-step examples, video tutorials with worked examples.

2. Completing the Square — video on Khan Academy

3. Completing the Square — video on Khan Academy

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Methods of Solving Quadratic Equations: Formula, Methods, Examples

In algebra, polynomials are algebraic expressions with exponents of the variables as whole numbers. When a polynomial is equated to zero, we get a polynomial equation. If a quadratic polynomial is equated to zero, then we can call it a quadratic equation. A quadratic equation is an equation of second degree.

A quadratic equation makes a \( \cup \)-shaped curve (parabola) if we represent it graphically. There are always two solutions to a quadratic equation with either real roots or imaginary roots. Let us learn in detail the different methods of solving quadratic equations.

Definition of Quadratic Equations

An equation of second-degree polynomial in one variable, such as \(x\) usually equated to zero, is called a quadratic equation. The coefficient of \({x^2}\) must not be zero in a quadratic equation. \(p(x)\) is a quadratic polynomial, then \(p(x) = 0\) called a quadratic equation. For example, \(3{x^2} + 2x + 2 = 0, – {x^2} + 6x + 1 = 0,7{x^2} – 6x + 4 = 0\)etc., are quadratic equations.

Standard Form of a Quadratic Equation    

The standard form of a quadratic equation is given by \(a{x^2} + bx + c = 0\) where\(a,b,c\) are real numbers, \(a \ne 0\) and \(a\) is the coefficient of \({x^2},b\) is the coefficient of \(x\) and \(c\) is a constant.

Roots of Quadratic Equations

The values of the variable, like \(x\) that satisfy the equation in one variable are called the roots of the equation. The roots of the quadratic equation may be real or imaginary.

If a quadratic polynomial is equated to zero, it becomes a quadratic equation. The values of \(x\) satisfying the equation are known as the roots of the quadratic equation. Note that the zeroes of the quadratic polynomial \(a{x^2} + bx + c\) and the roots of the quadratic equation \(a{x^2} + bx + c = 0\) are the same.

Methods to Solve the Quadratic Equations

There are some methods to solve the quadratic equation. They are,

1. Factorization method 2. Completing square method 3. Formula method 4. Graphical Method

Solving Quadratic Equation by Factorization Method

If we can factorize \(a{x^2} + bx + c,\,a \ne 0,\) into a product of two linear factors, then the roots of the quadratic equation \(a{x^2} + bx + c = 0\) can be found by equating each factor to zero.

Steps to Solve Quadratic Equation Using Factorization

If we can factorize \(a{x^2} + bx + c = 0,\,a \ne 0,\) into a product of two linear factors, then the roots of the quadratic equation \(a{x^2} + bx + c = 0\) can be found by equating each factor to zero.

Let us take an example and discuss it.

Consider a quadratic equation \({x^2} + 5x + 6 = 0.\)

1. Compare the given quadratic equation with the standard form \(a{x^2} + bx + c = 0\) and find the coefficients of \({x^2},x\) and the constant to get the values for \(a,b,c.\)

So, comparing \({x^2} + 5x + 6 = 0\) with \(a{x^2} + bx + c = 0\) we get, \(a = 1,b = 5,c = 6.\)

2. Now we will split \(b\) as the sum of two numbers such that the product of these two numbers \( = a \times c = ac\) We can factorize a quadratic equation when \(b\) can be split in \(v\) and \(w\) or \(b = v + w\) and \( = a\, \times c = ac.\)

So, \({x^2} + 5x + 6 = 0 \Rightarrow {x^2} + (2 + 3)x + 6 = 0 \Rightarrow {x^2} + 2x + 3x + 6 = 0.\)

3. Find the common factor of the first two and last two terms separately, such as the new two terms have the same common factor. So, \(x\) the common factor of the first two terms \({x^2} + 2x.\) Therefore, taking \(x\) outside, we get \({x^2} + 2x = x(x + 2).\)

\(3\) is the common factor of the last two terms \(3x + 6.\) Therefore, taking \(3\) outside, we get \(3x+6=3(x+2).\)

So, \({x^2} + 2x + 3x + 6 = 0\) can be written as \(x(x + 2) + 3(x + 2) = 0.\)

4. If we take out the same common factor, then we get the product of two linear polynomials. Here, we can see that \((x + 2)\) is the common factor in the new two terms. So, we can write \(x(x + 2) + 3(x + 2) = 0\,as\,(x + 2)(x + 3) = 0\) 5. At last, we will use the zero product rule to find the zeros. Zero product property says that when \(p \times q = 0\) then either \(p = 0\,or\,q = 0\) Therefore, \((x + 2) = 0,or(x + 3) = 0.\)

6. Hence, the solutions or roots of the quadratic equation \({x^2} + 5x + 6 = 0\) are \(x =  – 2,x =  – 3.\)

Solving Quadratic Equation by Completing the Square Method

A quadratic equation can be solved by the method of completing the square.

Steps to Solve Quadratic Equation by Completing the Square Method

Consider the quadratic equation, \(a{x^2} + bx + c = 0,a \ne 0\). Let us divide the equation by \(a\). \({x^2} + \frac{b}{a}x + \frac{c}{a} = 0\) Multiply and divide \(2\) to \(x\) term. \( \Rightarrow {x^2} + 2 \times \frac{b}{{2a}} \times x + \frac{c}{a} = 0\)

Now, to make the perfect square, we need to add and subtract \({\left( {\frac{b}{{2a}}} \right)^2}\) from L.H.S. \( \Rightarrow {x^2} + 2 \times \frac{b}{{2a}} \times x + {\left( {\frac{b}{{2a}}} \right)^2} – {\left( {\frac{b}{{2a}}} \right)^2} + \frac{c}{a} = 0\) \( \Rightarrow {\left( {x + \frac{b}{{2a}}} \right)^2} – {\left( {\frac{b}{{2a}}} \right)^2} + \frac{c}{a} = 0\)

Transferring \( – {\left( {\frac{b}{{2a}}} \right)^2} + \frac{c}{a}\) from \(L.H.S\) to \(R.H.S\) we have, \( \Rightarrow {\left( {x + \frac{b}{{2a}}} \right)^2} =  – \frac{c}{a} + {\left( {\frac{b}{{2a}}} \right)^2}\)

Taking square root on both sides we have, \( \Rightarrow x + \frac{b}{{2a}} =  \pm \sqrt { – \frac{c}{a} + {{\left( {\frac{b}{{2a}}} \right)}^2}} \) \( \Rightarrow x =  \pm \sqrt { – \frac{c}{a} + {{\left( {\frac{b}{{2a}}} \right)}^2}}  – \frac{b}{{2a}}\)

For example, consider the quadratic equation \(2{x^2} + 8x + 3 = 0\)

Let us divide the equation by \(a\, = \,2\) \(2\left( {{x^2} + \frac{8}{2}x + \frac{3}{2}} \right) = 0\) \( \Rightarrow {x^2} + 4x + \frac{3}{2} = 0\) \( \Rightarrow {x^2} + 2 \times 2x + \frac{3}{2} = 0\)

Now, to make the perfect square, we need to add and subtract \({(2)^2}\) from \(L.H.S.\) \( \Rightarrow {x^2} + 2 \times 2 \times x + {(2)^2} – {(2)^2} + \frac{3}{2} = 0\) \( \Rightarrow {(x + 2)^2} – {(2)^2} + \frac{3}{2} = 0\)

Transferring \( – {(2)^2} + \frac{3}{2}\) from \(L.H.S\,to\,R.H.S\) we have, \( \Rightarrow {(x + 2)^2} = {(2)^2} – \frac{3}{2}\)

Taking square root on both sides we have, \( \Rightarrow x + 2 =  \pm \sqrt {4 – \frac{3}{2}} \) \( \Rightarrow x =  \pm \sqrt {\frac{5}{2}}  – 2\)

Hence, the required solution of the quadratic equation \(2{x^2} + 8x + 3 = 0\) is \(x =  \pm \sqrt {\frac{5}{2}}  – 2\)

Solving Quadratic Equation by Formula Method

The roots of the quadratic equation \(a{x^2} + bx + c = 0\) are given by the quadratic formula

\(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)

For example, consider the quadratic equation \(3{x^2} – 5x + 2 = 0\)

From the given quadratic equation \(a = 3,\,b =  – 5,\,c = 2\)

The quadratic Equation formula is given by

\(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\) \(x = \frac{{ – ( – 5) \pm \sqrt {{{( – 5)}^2} – 4 \times 3 \times 2} }}{{2a}} = \frac{{ + 5 \pm \sqrt {25 – 24} }}{6}\) \( = x = \frac{{ – ( – 5) \pm \sqrt {{{( – 5)}^2} – 4 \times 3 \times 2} }}{{2 \times 3}} = \frac{{5 \pm \sqrt {25 – 24} }}{6}\) \( = \frac{{5 \pm \sqrt 1 }}{6} = \frac{{5 \pm 1}}{6} = \frac{{5 + 1}}{6},\frac{{5 – 1}}{6} = \frac{6}{6},\frac{4}{6}\) \( \Rightarrow x = 1\) or \(x = \frac{2}{3}\)

Hence, the roots of the given quadratic equation are \( \Rightarrow x = 1\) or \(x = \frac{2}{3}\)

Graphical Solution of Quadratic Equation

The standard form of a quadratic equation is \(a{x^2} + bx + c = 0,\) where \(a,b\) and \(c\) are real and \(a\,\, \ne \,\,0\)

Let us take an example and discuss \({x^2} – 5x + 6 = 0\) Considering \(p(x) = {x^2} – 5x + 6\) A quadratic polynomial always gives a parabolic graph. In a polynomial, the value of \(x\) which is responsible to make \(p(x) = 0\) is called the zero of the polynomial. When the polynomial equated with zero, it becomes an equation. If the zeros of the quadratic polynomial are known then they can be considered as the solutions of the equation which is formed by equating the polynomial to zero.

When \(x =  + 2,p(2) = {(2)^2} – 5(2) + 6 = 4 – 10 + 6 =  – 6 + 6 = 0\) \(x =  + 3,p(3) = {(3)^2} – 5(3) + 6 = 9 – 15 + 6 =  – 6 + 6 = 0\) Hence, the zeros of the quadratic polynomial are \( + 3,\, + 2\) If we represent the polynomial graphically then, the graph of the polynomial cuts the \(x – \)axis at \( + 3\,{\rm{and}}\, + 2\)

Let us take another example and discuss \({x^2} – 2x + 1 = 0\). Considering \(p(x) = {x^2} – 2x + 1\) A quadratic polynomial always gives a parabolic graph. When \(x =  + 1,p(1) = {(1)^2} – 2(1) + 1 = 1 – 2 + 1 =  – 1 + 1 = 0\) Hence, the zeros of the polynomial are \( + 1\) If we represent the polynomial graphically then, the graph of the polynomial touches the \(x – \)axis at \( + 1\). Therefore, the solution of the equation is \( + 1\).

Solved Examples

Q.1. Find the roots of the quadratic equation \(6{x^2} – x – 2 = 0\) Ans: We have \(6{x^2} – x – 2 = 0\) \( \Rightarrow 6{x^2} + 3x – 4x – 2 = 0\) \( \Rightarrow 3x(2x + 1) – 2(2x + 1) = 0\) The roots of \(6{x^2} – x – 2 = 0\) are the values of \(x\) for which \((3x – 2)(2x + 1) = 0\) Therefore, \(3x – 2 = 0\,\,or\,2x\,\, + \,\,1 = 0\) \(x = \frac{2}{3},x = \frac{{ – 1}}{2}\) Hence, the roots are \(\frac{2}{3}\& \frac{{ – 1}}{2}.\)

Q.2. Find the roots of the quadratic equation by using the formula method \(2{x^2} – 8x – 24 = 0\) Ans: From the given quadratic equation \(a = 2,b =  – 8,c =  – 24\) Quadratic equation formula is given by \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\) \(x = \frac{{ – ( – 8) \pm \sqrt {{{( – 8)}^2} – 4 \times 2 \times ( – 24)} }}{{2 \times 2}} = \frac{{8 \pm \sqrt {64 + 192} }}{4}\) \(x = \,\frac{{8 \pm \sqrt {256} }}{4}S = \frac{{8 \pm 16}}{4} = \frac{{8 + 16}}{4},\frac{{8 – 16}}{4} = \frac{{24}}{4},\frac{{ – 8}}{4}\) \( \Rightarrow x = 6,x =  – 2\) Hence, the roots of the given quadratic equation are \(6\&  – 2.\)

Q.3. Find the roots of the quadratic equation \({x^2} + 10x + 21 = 0\) by completing the square method. Ans: \({x^2} + 10x + 21 = 0\) \( \Rightarrow {x^2} + 10x =  – 21\) (Subtracted \(21\) from both sides of the equation) \( \Rightarrow {x^2} + 10x + 25 =  – 21 + 25\) (Added \({\left( {\frac{b}{2}} \right)^2} = {\left( {\frac{{10}}{2}} \right)^2} = 25\) on both the sides of the equation) \( \Rightarrow {(x + 5)^2} = 4\) (Completed the square by using the identity \({(a + b)^2} = {a^2} + 2ab + {b^2}\)) Then, take the square root on both sides. \(x + 5 =  \pm 2\) \(x =  – 3,x =  – 7\) Hence, the roots of the given quadratic equation are \( – 3\&  – 7.\)

Q.4. Find the roots of the quadratic equation\(2{x^2} + 8x + 3 = 0\) by completing the square method. Ans: \( \Rightarrow 2{x^2} + 8x =  – 3\) [Subtracted \(3\) from both sides of the equation] \( \Rightarrow {x^2} + 4x = \frac{{ – 3}}{2}\) (Divided both sides of the equation by \(2\)) \( \Rightarrow {x^2} + 4x+4 = \frac{{ – 3}}{2} + 4\) [Added \({\left( {\frac{b}{2}} \right)^2} = {\left( {\frac{4}{2}} \right)^2} = 4\) on both the sides of the equation] \(\Rightarrow(x+2)^{2}=\frac{5}{2}\) [Completed the square by using the identity \((a+b)^{2}=a^{2}+2 a b+b^{2}\)] Then, take the square root on both the sides \( \Rightarrow x + 2 =  \pm \sqrt {\frac{5}{2}} \) \( \Rightarrow x = -2 \pm \sqrt {\frac{5}{2}} \) is the required solutions.

Q.5. Find the roots of the quadratic equation \({x^2} + 3x – 10 = 0\) by factorization method. Ans: We have \({x^2} + 3x – 10 = 0\) \( \Rightarrow {x^2} + 5x – 2x – 10 = 0\) \( \Rightarrow x(x + 5) – 2(x + 5) = 0\) \( \Rightarrow (x – 2)(x + 5) = 0\) So, the roots of \({x^2} + 3x – 10 = 0\) are the values of \(x\) for which \((x – 2)(x + 5) = 0\) Therefore, \(x – 2 = 0\,\,or\,x + 5 = 0\) \(x = 2, – 5\,\) Hence, the roots are \(2\&  – 5.\)

In this article, we discussed quadratic equation in the variable \(x\) which is an equation of the form \(a{x^2} + bx + c = 0,\) where \(a,b,c\) are real numbers, \(a\, \ne 0\) Also, we discussed the methods of solving the quadratic equations such as factorizing method, completing the square method, formula method etc.

FAQs on Methods of Solving Quadratic Equations

Q.1. What are \(5\) methods of solving a quadratic equation? Ans: We can solve the quadratic equations by using different methods given below: 1. By factorizing method 2. By completing the square method 3. By using the quadratic formula 4. By using the graphical method 5. By using the trial and error method

Q.2. How can you identify a quadratic equation? Ans: An equation is a quadratic equation in the variable \(x\) if it is of the form \(a{x^2} + bx + c = 0\) where \(a,b,c\) are real numbers, \(a\, \ne \,0\).

Q.3. Which method can you use to solve all quadratic equations? Ans: We can not use factorizing method and completing square method for every quadratic equation as there are some constraints. We can use the formula method to solve all quadratic equations. The roots of the quadratic equation \(a{x^2} + bx + c = 0\) are given by: \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\) This is the quadratic formula for finding the roots of a quadratic equation.

Q.4. How many types of quadratic equations are there? Ans: There are three types of quadratic equations: 1. Standard form: \(a{x^2} + bx + c = 0,a \ne 0\) 2. Factored form: \((x – a)(x – b) = 0\) 3. Vertex form: \(a{(x – h)^2} + k = 0\) Each form of a quadratic equation has specific importance. Therefore, identifying the benefits of each different form can make it easier to understand and solve different situations.

Q.5. Can you use the quadratic formula for any quadratic equation? Ans: Yes, we can use the quadratic formula for any quadratic equation.

Q.6. What is the standard form of the quadratic equation? Ans: The form \(a{x^2} + bx + c = 0,\,a \ne 0\) is called the standard form of a quadratic equation.

We hope you find this detailed article on methods of solving quadratic equations helpful. If you have any doubts or queries regarding this topic, feel to ask us in the comment section.

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Quadratic Equations

An example of a Quadratic Equation :

The function can make nice curves like this one:

The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x 2 ).

It is also called an "Equation of Degree 2" (because of the "2" on the x )

Standard Form

The Standard Form of a Quadratic Equation looks like this:

• a , b and c are known values. a can't be 0
• x is the variable or unknown (we don't know it yet)

Here are some examples:

Have a Play With It

Play with the Quadratic Equation Explorer so you can see:

• the function's graph, and
• the solutions (called "roots").

Hidden Quadratic Equations!

As we saw before, the Standard Form of a Quadratic Equation is

But sometimes a quadratic equation does not look like that!

For example:

How To Solve Them?

The " solutions " to the Quadratic Equation are where it is equal to zero .

They are also called " roots ", or sometimes " zeros "

There are usually 2 solutions (as shown in this graph).

And there are a few different ways to find the solutions:

Just plug in the values of a, b and c, and do the calculations.

We will look at this method in more detail now.

About the Quadratic Formula

First of all what is that plus/minus thing that looks like ± ?

The ± means there are TWO answers:

x = −b + √(b 2 − 4ac) 2a

x = −b − √(b 2 − 4ac) 2a

Here is an example with two answers:

But it does not always work out like that!

• Imagine if the curve "just touches" the x-axis.
• Or imagine the curve is so high it doesn't even cross the x-axis!

This is where the "Discriminant" helps us ...

Discriminant

Do you see b 2 − 4ac in the formula above? It is called the Discriminant , because it can "discriminate" between the possible types of answer:

• when b 2 − 4ac is positive, we get two Real solutions
• when it is zero we get just ONE real solution (both answers are the same)
• when it is negative we get a pair of Complex solutions

Complex solutions? Let's talk about them after we see how to use the formula.

Using the Quadratic Formula

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.

Example: Solve 5x 2 + 6x + 1 = 0

Answer: x = −0.2 or x = −1

Let's check the answers:

Remembering The Formula

A kind reader suggested singing it to "Pop Goes the Weasel":

Try singing it a few times and it will get stuck in your head!

Or you can remember this story:

x = −b ± √(b 2 − 4ac) 2a

"A negative boy was thinking yes or no about going to a party, at the party he talked to a square boy but not to the 4 awesome chicks. It was all over at 2 am. "

Complex Solutions?

When the Discriminant (the value b 2 − 4ac ) is negative we get a pair of Complex solutions ... what does that mean?

It means our answer will include Imaginary Numbers . Wow!

Example: Solve 5x 2 + 2x + 1 = 0

Answer: x = −0.2 ± 0.4 i

The graph does not cross the x-axis. That is why we ended up with complex numbers.

In a way it is easier: we don't need more calculation, we leave it as −0.2 ± 0.4 i .

Example: Solve x 2 − 4x + 6.25 = 0

Answer: x = 2 ± 1.5 i

BUT an upside-down mirror image of our equation does cross the x-axis at 2 ± 1.5 (note: missing the i ).

Just an interesting fact for you!

• Quadratic Equation in Standard Form: ax 2 + bx + c = 0
• Quadratic Equations can be factored
• Quadratic Formula: x = −b ± √(b 2 − 4ac) 2a
• positive, there are 2 real solutions
• zero, there is one real solution
• negative, there are 2 complex solutions

The Quadratic Formula

What it is, what it does, and how to use it

What is the Quadratic Formula?

The quadratic formula is:

What does this formula tell us?

The quadratic formula calculates the solutions of any quadratic equation.

What is a quadratic equation?

A quadratic equation is an equation that can be written as ax ² + bx + c where a ≠ 0 . In other words, a quadratic equation must have a squared term as its highest power.

Examples of quadratic equations

$$ y = 5x^2 + 2x + 5 \\ y = 11x^2 + 22 \\ y = x^2 - 4x +5 \\ y = -x^2 + + 5 $$

Non Examples

$$ y = 11x + 22 \\ y = x^3 -x^2 +5x +5 \\ y = 2x^3 -4x^2 \\ y = -x^4 + 5 $$

Ok, but what is a 'solution'?

Well a solution can be thought in two ways:

Example of the quadratic formula to solve an equation

Use the formula to solve theQuadratic Equation: $$ y = x^2 + 2x + 1 $$ .

Just substitute a,b, and c into the general formula:

$$ a = 1 \\ b = 2 \\ c = 1 $$

Below is a picture representing the graph of y = x² + 2x + 1 and its solution.

Quadratic Formula Song

A catchy way to remember the quadratic formula is this song (pop goes the weasel).

Practice Problems

Use the quadratic formula to find the solutions to the following equation: y = x² − 2x + 1 and its solution.

• b = − 2

Use the quadratic formula to find the solutions to the following equation: y = x² − x − 2 and its solution .

In this quadratic equation, y = x² − x − 2 and its solution:

• b = − 1
• c = − 2

Use the quadratic formula to find the solutions to the following equation: y = x² − 1 and its solution.

In this quadratic equation, y = x² − 1 and its solution:

• c = −1

Calculate the solutions of the the quadratic equation below by using the quadratic formula : y = x² + 2x − 3 and its solution.

In this quadratic equation, y = x² + 2x − 3 and its solution:

• c = −3

Below is a picture of the graph of the quadratic equation and its two solutions.

Use the quadratic formula to find the solutions to the following equation: y = x² + 4x − 5 and its solution.

In this quadratic equation, y = x² + 4x − 5 and its solution:

• c = −5

Use the quadratic formula to find the solutions to the following equation: y = x² − 4x + 5 and its solution.

In this quadratic equation, y = x² − 4x + 5 and its solution:

• b = −4

Below is a picture of this quadratic's graph.

• Solving Quadratic Equations

• Discriminant
• Completing the Square in Math

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Solving Quadratics by Square Root Method

How to solve quadratic equations using the square root method.

This is the “best” method whenever the quadratic equation  only contains [latex]{x^2}[/latex] terms. That implies no presence of any [latex]x[/latex] term being raised to the first power somewhere in the equation.

The general approach is to collect all [latex]{x^2}[/latex] terms on one side of the equation while keeping the constants to the opposite side. After doing so, the next obvious step is to take the square roots of both sides to solve for the value of [latex]x[/latex]. Always attach the [latex] \pm [/latex] symbol when you get the square root of the constant.

Examples of How to Solve Quadratic Equations by Square Root Method

Example 1 : Solve the quadratic equation below using the Square Root Method.

I will isolate the only [latex]{x^2}[/latex] term on the left side by adding both sides by [latex] + 1[/latex]. Then solve the values of [latex]x[/latex] by taking the square roots of both sides of the equation. As I mentioned before, we need to attach the plus or minus symbol to the square root of the constant.

So I have [latex]x = 5[/latex] and [latex]x = – \,5[/latex] as final answers since both of these values satisfy the original quadratic equation. I will leave it to you to verify.

Example 2 : Solve the quadratic equation below using the Square Root Method.

This problem is very similar to the previous example. The only difference is that after I have separated the [latex]{x^2}[/latex] term and the constant in the opposite sides of the equation, I need to divide the equation by the coefficient of the squared term before taking the square roots of both sides.

The final answers are [latex]x = 4[/latex] and [latex]x = – \,4[/latex].

Example 3 : Solve the quadratic equation below using the Square Root Method.

I can see that I have two [latex]{x^2}[/latex] terms, one on each side of the equation. My approach is to collect all the squared terms of [latex]x[/latex] to the left side, and combine all the constants to the right side. Then solve for [latex]x[/latex] as usual, just like in Examples 1 and 2.

The solutions to this quadratic formula are [latex]x = 3[/latex] and [latex]x = – \,3[/latex].

Example 4 : Solve the quadratic equation below using the Square Root Method.

The two parentheses should not bother you at all. The fact remains that all variables come in the squared form, which is what we want. This problem is perfectly solvable using the square root method.

So my first step is to eliminate both of the parentheses by applying the distributive property of multiplication. Once they are gone, I can easily combine like terms. Keep the [latex]{x^2}[/latex] terms to the left, and constants to the right. Finally, apply square root operation in both sides and we’re done!

Not too bad, right?

Example 5 : Solve the quadratic equation below using the Square Root Method.

Since the [latex]x[/latex]-term is being raised to the second power twice, that means, I need to perform two square root operations in order to solve for [latex]x[/latex].

The first step is to have something like this: (   ) 2 = constant . This allows me to get rid of the exponent of the parenthesis on the first application of square root operation.

After doing so, what remains is the “stuff” inside the parenthesis which has an [latex]{x^2}[/latex] term. Well, this is great since I already know how to handle it just like the previous examples.

There’s an [latex]x[/latex]-squared term left after the first application of square root.

Now we have to break up [latex]{x^2} = \pm \,6 + 10[/latex] into two cases because of the “plus” or “minus” in [latex]6[/latex].

• Solve the first case where [latex]6[/latex] is positive .

• Solve the second case where [latex]6[/latex] is negative .

The solutions to this quadratic equations are [latex]x = 4[/latex], [latex]x = – \,4[/latex], [latex]x = 2[/latex], and [latex]x = – \,2[/latex]. Yes, we have four values of [latex]x[/latex] that can satisfy the original quadratic equation.

Example 6 : Solve the quadratic equation below using the Square Root Method.

Example 7 : Solve the quadratic equation below using the Square Root Method.

You may also be interested in these related math lessons or tutorials:

Solving Quadratic Equations by Factoring Method Solving Quadratic Equations by the Quadratic Formula Solving Quadratic Equations by Completing the Square

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Quadratic Formula : x = − b ± b 2 − 4 a c 2 a

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• Math Article

Quadratics or Quadratic Equations

Quadratics   can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations . The general form of the quadratic equation is: 

ax² + bx + c = 0

where x is an unknown variable and a, b, c are numerical coefficients. For example, x 2 + 2x +1 is a quadratic or quadratic equation. Here, a ≠ 0 because if it equals zero then the equation will not remain quadratic anymore and it will become a linear equation, such as: 

Thus, this equation cannot be called a quadratic equation.

The terms a, b and c are also called quadratic coefficients. 

The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations . The roots of any polynomial are the solutions for the given equation.

What is Quadratic Equation?

The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of:

where x is the unknown variable and a, b and c are the constant terms.

Standard Form of Quadratic Equation

Since the quadratic includes only one unknown term or variable, thus it is called univariate. The power of variable x is always non-negative integers. Hence the equation is a polynomial equation with the highest power as 2.

The solution for this equation is the values of x, which are also called zeros. Zeros of the polynomial are the solution for which the equation is satisfied. In the case of quadratics, there are two roots or zeros of the equation. And if we put the values of roots or x on the left-hand side of the equation, it will equal to zero. Therefore, they are called zeros.

Quadratics Formula

The formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the equation. Suppose ax² + bx + c = 0 is the quadratic equation, then the formula to find the roots of this equation will be:

x = [-b±√(b 2 -4ac)]/2a

The sign of plus/minus indicates there will be two solutions for x. Learn in detail the quadratic formula here.

Examples of Quadratics

Beneath are the illustrations of quadratic equations of the form (ax² + bx + c = 0)

• x² –x – 9 = 0
• 5x² – 2x – 6 = 0
• 3x² + 4x + 8 = 0
• -x² +6x + 12 = 0

Examples of a quadratic equation with the absence of a ‘ C ‘- a constant term.

• -x² – 9x = 0
• x² + 2x = 0
• -6x² – 3x = 0
• -5x² + x = 0
• -12x² + 13x = 0
• 11x² – 27x = 0

Following are the examples of a quadratic equation in factored form

• (x – 6)(x + 1) = 0  [ result obtained after solving is x² – 5x – 6 = 0]
• –3(x – 4)(2x + 3) = 0  [result obtained after solving is -6x² + 15x + 36 = 0]
• (x − 5)(x + 3) = 0  [result obtained after solving is x² − 2x − 15 = 0]
• (x – 5)(x + 2) = 0  [ result obtained after solving is x² – 3x – 10 = 0]
• (x – 4)(x + 2) = 0  [result obtained after solving is x² – 2x – 8 = 0]
• (2x+3)(3x – 2) = 0  [result obtained after solving is 6x² + 5x – 6]

Below are the examples of a quadratic equation with an absence of linear co – efficient ‘ bx’

• 2x² – 64 = 0
• x² – 16 = 0
• 9x² + 49 = 0
• -2x² – 4 = 0
• 4x² + 81 = 0
• -x² – 9 = 0

How to Solve Quadratic Equations?

There are basically four methods of solving quadratic equations. They are:

• Completing the square

Using Quadratic Formula

• Taking the square root

Factoring of Quadratics

• Begin with a equation of the form ax² + bx + c = 0
• Ensure that it is set to adequate zero.
• Factor the left-hand side of the equation by assuming zero on the right-hand side of the equation.
• Assign each factor equal to zero.
• Now solve the equation in order to determine the values of x.

Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors.

(2x+3)(x-2)=0

Learn more about the factorization of quadratic equations here.

Completing the Square Method

Let us learn this method with example.

Example: Solve 2x 2 – x – 1 = 0.

First, move the constant term to the other side of the equation.

2x 2 – x = 1

Dividing both sides by 2.

x 2 – x/2 = ½

Add the square of half of the coefficient of x, (b/2a) 2 , on both the sides, i.e., 1/16

x 2 – x/2 + 1/16 = ½ + 1/16

Now we can factor the right side,

(x-¼) 2 = 9/16 = (¾) 2

Taking root on both sides;

X – ¼ = ±3/4

Add ¼ on both sides

X = ¼ + ¾ = 4/4 = 1

X = ¼ – ¾ = -2/4 = -½ 

To learn more about completing the square method, click here .

For the given Quadratic equation of the form, ax² + bx + c = 0

Therefore the roots of the given equation can be found by:

\(\begin{array}{l}x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\end{array} \)

where ± (one plus and one minus) represent two distinct roots of the given equation.

Taking the Square Root

We can use this method for the equations such as:

x 2 + a 2 = 0

Example: Solve x 2 – 50 = 0.

x 2 – 50 = 0

Taking the roots both sides

√x 2 = ±√50

x = ±√(2 x 5 x 5)

Thus, we got the required solution.

Related Articles

Video Lesson on Quadratic Equations

Range of quadratic equations.

Solved Problems on Quadratic Equations

Applications of Quadratic Equations

Many real-life word problems can be solved using quadratic equations. While solving word problems, some common quadratic equation applications include speed problems and Geometry area problems.

• Solving the problems related to finding the area of quadrilateral such as rectangle, parallelogram and so on
• Solving Word Problems involving Distance, speed, and time, etc.,

Example: Find the width of a rectangle of area 336 cm2 if its length is equal to the 4 more than twice its width. Solution: Let x cm be the width of the rectangle. Length = (2x + 4) cm We know that Area of rectangle = Length x Width x(2x + 4) = 336 2x 2 + 4x – 336 = 0 x 2 + 2x – 168 = 0 x 2 + 14x – 12x – 168 = 0 x(x + 14) – 12(x + 14) = 0 (x + 14)(x – 12) = 0 x = -14, x = 12 Measurement cannot be negative. Therefore, Width of the rectangle = x = 12 cm

Practice Questions

• Solve x 2 + 2 x + 1 = 0.
• Solve 5x 2 + 6x + 1 = 0
• Solve 2x 2 + 3 x + 2 = 0.
• Solve x 2 − 4x + 6.25 = 0

Frequently Asked Questions on Quadratics

What is a quadratic equation, what are the methods to solve a quadratic equation, is x 2 – 1 a quadratic equation, what is the solution of x 2 + 4 = 0, write the quadratic equation in the form of sum and product of roots..

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Thanks a lot ,This was very useful for me

x=√9 Squaring both the sides, x^2 = 9 x^2 – 9 = 0 It is a quadratic equation.

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