write an flowchart to find the roots of a quadratic equation

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Quadratic Equation Algorithm Flowchart

Roots of Quadratic Equation

A quadratic equation is an algebraic equation whose degree is two. We shall learn how to find the roots of quadratic equations algebraically and using the quadratic formula. The general form of a quadratic equation is ax 2 + bx + c = 0, where x is the unknown and a, b and c are known quantities such that a ≠ 0. Since the degree of such an equation is two, we get two roots of any given quadratic equation.

Learn more about quadratic equations .

In this article, we shall learn and practice solving a quadratic equation by factoring the given equation and using the quadratic formula. Also, learn how to find the nature of the roots of a quadratic equation.

What is meant by the Roots of a Quadratic Equation?

Roots of a quadratic equation mean those values of the unknown for which the given equation equals zero. Let ax 2 + bx + c = 0 is a quadratic equation (a ≠ 0), if for x = 𝛼 and x = 𝛽 the given function vanishes. We say 𝛼 and 𝛽 are the roots of the quadratic equation and (x – 𝛼) and (x – 𝛽) are the factors of the quadratic equation.

Graphically, the roots of a quadratic equation mean those points where the curve of the given quadratic equation intersects the x-axis.

How to Find Roots of Quadratic Equation

Now, we shall discuss how to find the roots of a quadratic equation by different methods.

Completing the Squares Method

We shall reduce the given quadratic equation into any of the standard algebraic identities and then find the root of the equation. Let us take an example 2x 2 – 5x – 3 = 0

to understand this method.

2x 2 – 5x – 3 = 0

⇒ x 2 – (5/2)x – 3/2 = 0 (Dividing both sides by coefficient of x 2 )

⇒ x 2 – 2. ½ . 5/2. x – 3/2 = 0

[adjusting the given quadratic equation to reduce it into the form (a – b) 2 ]

⇒ x 2 – (2.5)/(4.x) – 3/2 = 0

⇒ x 2 – (2.5)/(4.x) + 25/16 – 3/2 = 25/16 (adding both sides by 25/16)

⇒ x 2 – 2.(5/4).x + (5/4) 2 = 25/16 + 3/2

⇒ (x – 5/4) 2 = 49/16

⇒ x – 5/4 = ± 7/4

⇒ x = 5/4 ± 7/4

∴ x = 5/4 + 7/4 and x = 5/4 – 7/4

⇒ x = 3 and x = –½

Split the Middle Term Method

In this method we find the roots of the given quadratic equation ax 2 + bx + c = 0 by splitting the coefficient of x. We have to find two numbers 𝛼 and 𝛽 such that 𝛼 + 𝛽 = b and 𝛼𝛽 = c. For example, we have to find the roots of x 2 – 5x + 6 = 0

• 6 = 2 × 3 and –5 = –3 –2
• We shall split the middle term as: x 2 –3x –2x + 6 = 0
• Factorising: x(x –3) –2 (x – 3) = 0 ⇒ (x –2)(x – 3) = 0
• ∴ x = 2 and 3

Using Quadratic Equation

For any given quadratic equation ax 2 + bx + c = 0, the roots of the equation are given by the following formula:

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

For example, for the quadratic equation x 2 – 5x + 3 = 0

\(\begin{array}{l}x=\frac{-(-5)\pm \sqrt{(-5)^{2}-4\times 1\times 3}}{2\times 1}\end{array} \)

\(\begin{array}{l}\Rightarrow x=\frac{5\pm \sqrt{25-12}}{2}= \frac{5 \pm \sqrt{13}}{2}\end{array} \)

∴ x = [5 + √13]/2 and [5 – √13]/2

Nature of Roots of Quadratic Equation

The nature of the roots of a quadratic equation is determined by finding the value of the discriminant D.

For the quadratic equation ax 2 + bx + c = 0; a ≠ 0

Discriminant D = b 2 – 4ac

Learn more about nature of roots of quadratic equation .

Video Lessons

Visualising square roots.

Finding Square roots

Related Articles

• Quadratic Equations for Class 10
• Quadratic Formula
• Factorisation
• Polynomials

Solved Examples on Roots of Quadratic Equations

Find the roots of the following quadratic equation using completing the square method:

x 2 – 3x – 10 = 0.

We have x 2 – 3x – 10 = 0

Multiplying and dividing the middle term by 2, we get

⇒ x 2 – 2. ½. 3x – 10 = 0

Adding both sides by 9/4, we get

⇒ x 2 – 2.(3/2).x – 10 + 9/4 = 9/4

⇒ x 2 – 2.(3/2).x + 9/4 = 10 + 9/4

⇒ (x – 3/2) 2 = 49/4 {∴ (a + b) 2 = a 2 + 2ab + b 2 }

Taking square roots on both the sides, we get

⇒ x – 3/2 = ±7/2

⇒ x = 7/2 + 3/2 and –7/2 + 3/2

⇒ x = 5 and –2

Find the roots of the quadratic equation: x 2 – 8x + 15 = 0

We shall use the quadratic formula,

\(\begin{array}{l}\Rightarrow x=\frac{-(-8)\pm \sqrt{(-8)^{2}-4.1.15}}{2\times 1}\end{array} \)

\(\begin{array}{l}\Rightarrow x=\frac{8\pm \sqrt{64-60}}{2}=\frac{8\pm \sqrt{4}}{2}\end{array} \)

⇒ x = (8 + 2)/2 and (8 – 2)/2

⇒ x = 5 and 3.

Frequently Asked Questions on Roots of Quadratic Equation

What is meant by the root of a quadratic equation.

The roots of a quadratic equation are those values of the unknown for the equation is equal to zero.

How many roots does a quadratic equation have?

A quadratic equation has 2 roots, they may be equal or distinct.

How to find the roots of a quadratic equation?

The roots of a quadratic equation may be determined by factorising the given equation and by using the quadratic formula.

How to determine whether the given quadratic equation has real roots?

If the given quadratic equation has real roots then the value of the discriminant D ≥ 0.

What is the condition for getting equal roots for a quadratic equation?

If the value of the discriminant D = 0 then the quadratic equation has real roots.

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Roots of Quadratic Equation

For a given quadratic equation ax 2 + bx + c = 0, the values of x that satisfy the equation are known as its roots. i.e., they are the values of the variable (x) which satisfies the equation. The roots of a quadratic function are the x-coordinates of the x-intercepts of the function. Since the degree of a quadratic equation is 2, it can have a maximum of 2 roots. We can find the roots of quadratic equations using different methods.

• Factoring (when possible)
• Quadratic Formula
• Completing the Square
• Graphing (used to find only real roots)

Let us understand more about the roots of the quadratic equation along with discriminant, nature of the roots, the sum of roots, the product of roots, and more along with some examples.

The roots of a quadratic equation are the values of the variable that satisfy the equation. They are also known as the "solutions" or "zeros" of the quadratic equation . For example, the roots of the quadratic equation x 2 - 7x + 10 = 0 are x = 2 and x = 5 because they satisfy the equation. i.e., when each of them is substituted in the given equation we get 0.

• when x = 2, 2 2 - 7(2) + 10 = 4 - 14 + 10 = 0.
• when x = 5, 5 2 - 7(5) + 10 = 25 - 35 + 10 = 0.

But how to find the roots of a general quadratic equation ax 2 + bx + c = 0? Let us try to solve it for x by completing the square.

ax 2 + bx = - c

Dividing both sides by 'a',

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

Here, the coefficient of x is b/a. Half of it is b/(2a). Its square is b 2 /4a 2 . Adding b 2 /4a 2 on both sides,

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

[ x + (b/2a) ] 2 = (b 2 - 4ac) / 4a 2 (using (a + b)^2 formula )

Taking square root on both sides,

x + (b/2a) = ±√ [(b 2 - 4ac) / 4a 2 ]

x + (b/2a) = ±√ (b 2 - 4ac) / 2a

Subtracting b/2a from both sides,

x = - (b/2a) ±√ (b 2 - 4ac) / 2a

x = (-b ± √ (b 2 - 4ac) )/2a

This is known as the quadratic formula and it can be used to find any type of roots of a quadratic equation.

How to Find the Roots of Quadratic Equation?

The process of finding the roots of the quadratic equations is known as "solving quadratic equations". In the previous section, we have seen that the roots of a quadratic equation can be found using the quadratic formula. Along with this method, we have several other methods to find the roots of a quadratic equation. To know about these methods in detail, click here . Let us discuss each of these methods here by solving an example of finding the roots of the quadratic equation x 2 - 7x + 10 = 0 (which was mentioned in the previous section) in each case. Note that, in each of these methods, the equation should be in the standard form ax 2 + bx + c = 0.

Finding Roots of Quadratic Equation by Factoring

• Factor the left side part.
• Set each of these factors to zero and solve.

Example: Find the roots of the quadratic equation x 2 - 7x + 10 = 0 by factoring.

By factoring the quadratic expression, we get (x - 2) (x - 5) = 0.

Now, setting each factor to zero and solving, we get

x - 2 = 0, x - 5 = 0

x = 2, x = 5.

Finding Roots of Quadratic Equation by Quadratic Formula

• Find a, b, and c values by comparing the given equation with ax 2 + bx + c = 0.
• Substitute them in the quadratic formula and simplify.

Example: Find the roots of quadratic equation x 2 - 7x + 10 = 0 using quadratic formula.

Here, a = 1, b = -7 and c = 10. Then by quadratic formula:

x = [-(-7) ± √((-7) 2 - 4(1)(10))] / (2(1))

= [ 7 ± √(49 - 40) ] / 2

= [ 7 ± √(9) ] / 2

= [ 7 ± 3 ] / 2

= (7 + 3) / 2, (7 - 3) / 2

= 10/2, 4/2

Therefore, x = 2, x = 5.

Finding Roots of Quadratic Equation by Completing Square

• Complete the square on the left side.
• Solve by taking square root on both sides.

Example: Find the quadratic roots of x 2 - 7x + 10 = 0 by completing square.

By completing the square, we get (x - (7/2) ) 2 = 9/4.

Now, taking the square root on both sides:

x - 7/2 = ± 3/2

x - 7/2 = 3/2, x - 7/2 = -3/2

x = 10/2, x = 4/2

x = 5, x = 2

Finding Roots of Quadratic Equation by Graphing

• Graph the left side part (the quadratic function ) either manually or using the graphing display calculator (GDC).
• Identify the x-intercepts which are nothing but the roots of the quadratic equation.

Example: Find the roots of the quadratic equation x 2 - 7x + 10 = 0 by graphing.

Solution: To solve this, we just need to graph f(x) = x 2 - 7x + 10 and identify the x-intercepts.

Therefore, the roots of the quadratic equation are x = 2 and x = 5 .

We can observe that the roots of the quadratic equation x 2 - 7x + 10 = 0 are x = 2 and x = 5 in each of the methods. Note that the factoring method works only when the quadratic equation is factorable; and we cannot find the complex roots of the quadratic equation using the graphing method. So the best methods that always work for finding the roots are the quadratic root formula and completing the square methods.

Nature of Roots of Quadratic Equation

The nature of the roots of a quadratic equation talks about "how many roots the equation has?" and "what type of roots the equation has?". A quadratic equation can have:

• two real and different roots
• two complex roots
• two real and equal roots (it means only one real root)

For example, in the above example, the roots of the quadratic equation x 2 - 7x + 10 = 0 are x = 2 and x = 5, where both 2 and 5 are two different real numbers , and so we can say that the equation has two real and different roots. But for finding the nature of the roots, we don't actually need to solve the equation. We can determine the nature of the roots by using the discriminant . The discriminant of the quadratic equation ax 2 + bx + c = 0 is D = b 2 - 4ac .

The roots of quadratic equation formula is x = (-b ± √ (b 2 - 4ac) )/2a. So this can be written as x = (-b ± √ D )/2a. Since the discriminant D is in the square root , we can determine the nature of the roots depending on whether D is positive, negative, or zero.

Nature of Roots When D > 0

Then the above formula becomes,

x = (-b ± √ positive number )/2a and it gives us two real and different roots. Thus, the quadratic equation has two real and different roots when b 2 - 4ac > 0.

Nature of Roots When D < 0

x = (-b ± √ negative number )/2a and it gives us two complex roots (which are different) as the square root of a negative number is a complex number . Thus, the quadratic equation has two complex roots when b 2 - 4ac < 0.

Note: A quadratic equation can never have one complex root. The complex roots always occur in pairs. i.e., if a + bi is a root then a - bi is also a root.

Nature of Roots When D = 0

x = (-b ± √ 0 )/2a = -b/2a

and hence the equation has only one real root. Thus, the quadratic equation has only one real root (or two equal roots -b/2a and -b/2a) when b 2 - 4ac = 0.

Sum and Product of Roots of Quadratic Equation

We have seen that the roots of the quadratic equation x 2 - 7x + 10 = 0 are x = 2 and x = 5. So the sum of its roots = 2 + 5 = 7 and the product of its roots = 2 × 5 = 10. But the sum and the product of roots of a quadratic equation ax 2 + bx + c = 0 can be found without actually calculating the roots. Let us see how.

We know that the roots of an equation ax 2 + bx + c = 0 by quadratic formula are (-b + √ (b 2 - 4ac)) /2a and (-b - √ (b 2 - 4ac) )/2a. Let us represent these by x 1 and x 2 respectively.

Sum of Roots of Quadratic Equation

The sum of the roots = x 1 + x 2

= (-b + √ (b 2 - 4ac)) /2a + (-b - √ (b 2 - 4ac) )/2a

= -b/2a - b/2a

Therefore, the sum of the roots of the quadratic equation ax 2 + bx + c = 0 is -b/a.

For the equation, x 2 - 7x + 10 = 0, the sum of the roots = -(-7)/1 = 7 (which was the sum of the actual roots 2 and 5).

Product of Roots of Quadratic Equation

The product of the roots = x 1 · x 2

= (-b + √ (b² - 4ac) )/2a · (-b - √ (b² - 4ac) )/2a

= (-b/2a) 2 - ( √ (b 2 - 4ac)/ 2a) 2 ( by a² - b² formula )

= b 2 / 4a 2 - (b 2 - 4ac) / 4a 2

= (b 2 / 4a 2 ) - (b 2 / 4a 2 ) + (4ac / 4a 2 )

= 4ac / 4a 2

Therefore, the product of the roots of the quadratic equation ax 2 + bx + c = 0 is c/a.

For the equation, x 2 - 7x + 10 = 0, the product of the roots = 10/1 = 10 (which was the product of the actual roots 2 and 5).

Important Formulas on Roots of Quadratic Equations:

For a quadratic equation ax 2 + bx + c = 0,

• The roots are calculated using the formula, x = (-b ± √ (b 2 - 4ac) )/2a.
• Discriminant is, D = b 2 - 4ac. If D > 0, then the equation has two real and distinct roots. If D < 0, the equation has two complex roots. If D = 0, the equation has only one real root.
• Sum of the roots = -b/a
• Product of the roots = c/a

☛ Related Topics:

• Roots of Quadratic Equation Calculator
• Quadratic Root Formula Calculator
• Quadratic Roots by Completing Square Calculator

Examples on Roots of Quadratic Equation

Example 1: Find the roots of the quadratic equation √2 p 2 + 7p + 5√2 = 0.

Let us find the roots using the factoring method.

Comparing the given equation with ap 2 + bp + c = 0, a = √2, b = 7 and c = 5√2.

Here, ac = (√2) (5√2) = 10 and b = 7.

Two numbers whose sum is 7 and whose product is 10 are 2 and 5. Let us split the middle term using these numbers.

√2 p 2 + 2p + 5p + 5√2 = 0

√2 p (p + √2) + 5 (p + √2) = 0

(p + √2) (√2 p + 5) = 0

p + √2 = 0; √2 p + 5 = 0

p = -√2; p = -5/√2 (or) -5√2/2

Answer: The roots of the given quadratic equation are -√2 and -5√2 / 2.

Example 2: Find the value(s) of k if the quadratic equation 3x 2 + kx + 2 = 0 has equal roots.

The quadratic equation ax 2 + bx + c = 0 has equal roots if its discriminant b 2 - 4ac = 0.

Here, a = 3, b = k, and c = 2.

b 2 - 4ac = 0

k 2 - 4(3)(2) = 0

k 2 - 24 = 0

Answer: When the given quadratic equation has equal roots, k = 2√6 or k = -2√6.

Example 3: Find the sum and the product of the roots of the equation (p + 1) x 2 - 2px + (q + 1) = 0 in terms p and q.

Comparing the given equation with ax 2 + bx + c = 0,

a = p + 1, b = -2p, and c = q + 1.

The sum of the roots = -b/a = -(-2p) / (p + 1) = 2p / (p + 1).

The product of the roots = c/a = (q + 1) / (p + 1).

Answer: The sum of the roots is 2p / (p + 1) and the product of the roots is (q + 1) / (p + 1).

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Practice Questions on Roots of Quadratic Equation

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FAQs on Roots of Quadratic Equation

What are the roots of a quadratic equation.

The roots of a quadratic equation ax 2 + bx + c = 0 are the values of the variable (x) that satisfy the equation. For example, the roots of the equation x 2 + 5x + 6 = 0 are -2 and -3. The roots of an equation ax 2 + bx + c = 0 can be found by the quadratic formula x = (-b ± √ (b 2 - 4ac)) /2a.

How Can We Find the Roots of Quadratic Equation?

The roots of a quadratic equation ax 2 + bx + c = 0 can be found using the quadratic formula that says x = (-b ± √ (b 2 - 4ac)) /2a. Alternatively, if the quadratic expression is factorable, then we can factor it and set the factors to zero to find the roots.

What are the Three Types of Roots of Roots of Quadratic Equation?

A quadratic equation ax 2 + bx + c = 0 can have:

• two real and distinct roots when b 2 - 4ac > 0.
• two complex roots when b 2 - 4ac < 0.
• two real and equal roots when b 2 - 4ac = 0.

How to Find the Roots of Quadratic Equation by Completing Square?

To find the roots of a quadratic equation ax 2 + bx + c = 0 by completing square, complete the square on the left side first. Then solve for x by taking the square root on both sides.

Where to Find Quadratic Root calculator?

The quadratic root calculator can be found by clicking here . This allows us to enter a quadratic equation and then it shows the roots along with step-by-step calculation.

How to Determine the Nature of Roots of Quadratic Equation?

The nature of the roots of an equation ax 2 + bx + c = 0 is determined by its discriminant, D = b 2 - 4ac.

• If D > 0, the equation has two real and distinct roots.
• If D < 0, the equation has two complex roots.
• If D = 0, the equation has two equal real roots.

How to Find the Roots of Quadratic Equation Using Quadratic Formula?

The quadratic formula says the roots of a quadratic equation ax 2 + bx + c = 0 are given by x = (-b ± √ (b 2 - 4ac)) /2a. To solve any quadratic equation, convert it into standard form ax 2 + bx + c = 0, find the values of a, b, and c, substitute them in the roots of quadratic equation formula and simplify.

How to Find the Sum and Product of Roots of Quadratic Equation?

For any quadratic equation ax 2 + bx + c = 0 whose roots are α and β,

• the sum of the roots, α + β = -b/a
• the product of the roots, α × β = c/a

Can Both the Roots of Quadratic Equation be Zeros?

Yes, both the roots of a quadratic equation can be zeros . For example, the two roots of the quadratic equation x 2 = 0 are 0 and 0.

How to Find the Roots of Quadratic Equation by Factoring?

To find the roots of an equation ax 2 + bx + c = 0 by factoring, factor its left side part, set each of the factors to zero and solve.

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• What are complex roots?
• Complex roots are the imaginary roots of a function.
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• To find the complex roots of a quadratic equation use the formula: x = (-b±i√(4ac – b2))/2a

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Quadratic equations are the equations where a polynomial has degree two. Roots of Quadratic Equations are the values of the variable that satisfies the equation.

Let’s learn how to find the roots of quadratic equation formulas with the help of solved examples.

Table of Content

• What is Roots of a Quadratic Equation?

How to Find Roots of Quadratic Equation?

• Solved Problems
• Practice Problems

Roots of a Quadratic Equation are the values of the variable let’s say x for which the equation gets satisfied.

• Roots of Quadratic Equations are also called Zeros of a Quadratic Equation or Solutions of a Quadratic Equation.
• Quadratic equations are mathematical expressions of the form ax 2 + bx + c = 0, where a, b, and c are constants, and x represents the variable.
• Solving for the values of x that make this equation true yields the roots of the quadratic equation.

Methods to Find Roots

The methods to find Roots or Zeroes of a Quadratic Equation are :

• Completing the Square
• Graphical Method

To Find roots of Quadratic Equations follow the methods mentioned below:

Finding Roots of Quadratic Equation by Quadratic Formula

x = [-b±√(b 2 – 4ac)]/2a

Example: The length of sides of a rectangle is given by x – 3 and x – 5 and the area of the rectangle is 3 unit 2 . Find the sides of the rectangle.

Area of rectangle = length*breadth = (x – 3)(x – 5) = 3 Area = x 2 – 8x + 15 = 3         = x 2 – 8x + 12 = 0 Discriminant = b 2 – 4ac = 64 – (4(1)(12)) = 64 – 48 = 16  x = [-b ± √(b 2 -4ac)]/2a = [-(-8) ± √16]/2 = (8±4)/2  x = 12/2 or 4/2 x = 2 or 6 When x is 2, sides are x – 3 = 2 – 3 = -1 and x – 5 = 2 – 5 = -3. Since length of sides cannot be equal therefore x = 2 is not a valid ans. When x is 6, sides are x – 3 = 6 – 3 = 3 and x – 5= 6 – 5 =1. Therefore, x = 6 is the valid answer and the sides are 3 and 1.

Finding Roots of Quadratic Equation by Factoring

A quadratic equation can be considered a factor of two terms. Like ax 2 + bx + c = 0 can be written as (x – x 1 )(x – x 2 ) = 0 where x 1 and x 2 are roots of quadratic equation.

• Find two numbers such that there product = ac and their sum = b.
• Then write x coefficient as sum of these two numbers and split them such that you get two terms for x.
• factor the first two as a group and last two terms as a group.
• Take common factors from these and on equating the two expression with zero after taking common factors and rearranging the equation we get the roots .

Example: Let be the quadratic equation x 2 + 3x = 18

x 2 + 3x – 18 = 0 Step: 1. 6 and -3 are the numbers whose sum is equal to b and product is equal ac. 2. x 2 + (6-3)x – 18 = x 2 + 6x -3x – 18 =0   3. x(x + 6) – x(x + 6) = 0 4. taking (x + 6) as common.  (x + 6)(x – 3) = 0 x = -6 or x = 3

In a factorizing method it is not necessary that you will always find these two numbers easily(especially in the case when roots are imaginary or irrational) so it is better to use the quadratic formula.

Nature of Roots

The nature of roots depends on the discriminant of the quadratic equation. The discriminant of a quadratic equation is given by b 2 – 4ac.  It is so because in quadratic formula square root of discriminant is there.

Root 1: If b 2 – 4ac > 0 roots are real and different. As the discriminant is >0 then the square root of it will not be imaginary. It has two cases.

• If b 2 – 4ac is a perfect square then roots are rational. As the discriminant is a perfect square, so we will have an integer as a square root of the discriminant. Hence, the roots are rational numbers.

Example: Let the quadratic equation be x 2 -5x+6=0.

Then the discriminant of the given equation is b 2 – 4ac=(-5) 2 – 4×1×6 = 25-24 = 1 According to Shridharacharaya formula    x = [-b±√(b 2 -4ac)]/2a = x = [-(-5) ± √1]/2     x 1 = [-(-5) + √1]/2 = 6/2 = 3    x 2 = [-(-5) – √1]/2 = 4/2 = 2    Therefore, the roots are 3,2. Both are rational and different.

• If b 2 – 4ac is not a perfect square then the square root of discriminant is irrational hence roots are irrational and occurs in pair.

Example: Let the quadratic equation be x 2 -7x+8 = 0.

Then the discriminant of the given equation is b 2 – 4ac=(-7) 2 – 4*1*8 = 49-32 = 17 According to Shridharacharaya formula x = [-b±√(b 2 -4ac)]/2a =  x = [-(-7) ± √17]/2  x 1 = [-(-7) + √17]/2 = [7 + √17]/2 x 2 = [-(-7) – √17]/2 = [7 – √17]/2 Therefore, the roots are [7 + √17]/2,[7 – √17]/2. Both are irrational and in pairs.

Root 2: If b 2 – 4ac = 0 roots are real and equal.

Example: Let the quadratic equation be 3x 2 -6x+3=0.

Then the discriminant of the given equation is b 2 – 4ac=(-6) 2 – 4*3*3 = 36 – 36 = 0 According to Shridharacharaya formula x = [-b±√(b 2 -4ac)]/2a =  x = [-(-6) ± √0]/[(2)(3)] x 1 = [-(-6) + √0]/2 = 6/6 = 1 x 2 = [-(-6) – √0]/2 = 6/6 = 1 Therefore, the roots are 1,1. Both are real and equal.

Root 3: If b 2 – 4ac < 0 roots are imaginary, or you can say complex roots. It is imaginary because the term under the square root is negative. These complex roots will always occur in pairs i.e, both the roots are conjugate of each other.

Example: Let the quadratic equation be x 2 +6x+11=0.

Then the discriminant of the given equation is b 2 – 4ac=(6) 2 – 4*1*11 = 36-44 = -8 According to Shridharacharaya formula x = [-b±√(b 2 -4ac)]/2a = x = [-(6) ± √(-8)]/2  x 1 = [-(6) + √(-8)]/2 = [-6 + √8i]/2 = 2[-3 + √2i]/2 = -3 + √2i x 2 = [-(6) – √(-8)]/2 = [-6 – √8i]/2 = 2[-3 – √2i]/2 = -3 – √2i Therefore, the roots are 3,2. Both are imaginary and conjugate of each other(in pair).

Learn More :

• Nature of Roots of Quadratic Equations

Graphs for Roots

The maximum/minimum value of quadratic function is found at x = -b/2a

We get maxima or minima when d(f(x))/dx = 0.

On differentiating quadratic function f(x) = ax 2 + bx + c. 

2ax + b = 0 

This x is either maxima(a<0) or minima(a>0). 

1. When D > 0

Graph of b 2 – 4ac > 0 is shown below:

2. When D = 0

Graph of b 2 – 4ac = 0 is shon below:

3. When D < 0

Graph of b 2 – 4ac < 0 is shown below:

• Quadratic Equations

Examples on Roots of Quadratic Equation

Question 1. The height of a triangle is less than 4 cm than the base. The area of triangle is 30 cm 2 . Find the height and base of triangle.

Let the base of triangle be x cm then height is x-4 cm  Area of triangle = 1/2*height*base = 1/2*(x)(x – 4)=30 Area = x 2 – 4x  = 30*2         = x 2 – 4x = 60         = x 2 – 4x – 60 = 0 Discriminant = (-4) 2 – 4(1)(-60) = 16+240 = 256 x = [-b±√(b 2 – 4ac)]/2a = [-(-4)±√256]/2 = (4±16)/2 x = 20/2 or -12/2 x = 10 or -6 As side cannot be negative, therefore -6 is not correct. So, when x is 10, base =10 cm and height =x – 4 = 10 – 4 = 6 cm Therefore, x = 10 is the valid answer and the base and height of the triangle are 10 and 6 respectively.

Question 2. The volume of a box is 600 inch 2 . The length of box is 2 inches less than the width. The height of box is 5 inches . Find the dimensions of box.

Let the width of box be x inches then length = x – 2 inches. Volume of box =Length* Width *Height = (x-2)(x)5= 600 x 2 – 2x = 120 => x 2 – 2x -120 = 0 x 2 + 10x – 12x – 120 = 0 x(x + 10) – 12(x + 10)=0 (x – 12)(x + 10)=0 x = 12 or x = -10 As width can not be negative, therefore, -10 is not correct. When x = 12, width = 12 inches, length = x – 2 = 12 – 2 = 10 inches, height = 5 inches 

Question 3. A ball is thrown from the top of a building . its height in meters above the ground as a function of time is given by h(t) = -4t 2 + 24t + 3. a) How much time it take to reach the maximum height and what is the maximum height. b) Find also the time at which ball hits the ground.

a) Since a<0, therefore the time to reach maximum height is = -b/2a. (refer graph part) t = -24/(2(-4)) = -24/-8 t = 3 sec. Height = h(t) = -4(3) 2 +24(3)+3 = 39 meters. b) When the ball hits the ground h(t)=0. -4t 2 + 24t + 3 = 0 The discriminant of the given equation is b 2 – 4ac=(24) 2 – 4×(-4)×3 = 576 + 48 = 624 According to Quadratic formula x = [-b±√(b 2 -4ac)]/2a = x = [-(24) ± √624]/[(2)(-4)] x 1 = [-24 + √624]/-8= 4 [-6 + √39]/-8 = [-6 + √39]/-2 = [6 – √39]/2 = -0.122499 = -0.1225(approx) x 2 = [-24 – √624]/-8= 4 [-6 – √39]/-8 = [-6 – √39]/-2 = [6 + √39]/2 = 6.122499 = 6.1225(approx) As time can not be negative, therefore, [6 – √39]/2 sec is not correct. Therefore, the ball hits the ground at [6 + √39]/2 = 6.122499 = 6.1225 sec

Practice Problems on Roots of Quadratic Equations

Problem 1: Find the roots of the following Quadratic Equations:

• 3x 2 – 7x + 2 = 0
• x 2 + 4x + 4 = 0
• 2x 2 + 5x – 3 = 0
• 4x 2 – 12x + 9 = 0
• x 2 – 6x + 9 = 0

Problem 2: The sum of two consecutive integers is 31. Find the two integers.

Problem 3: A quadratic equation of the form ax 2 + bx + c = 0 has roots x = 3 and x = -2. Find the values of a, b and c.

Problem 4: A ball is thrown into the air from a height of 5 feet with an initial velocity of 40 feet per second. The height h of the ball above the ground after t seconds can be modelled by the quadratic equation h(t) = -16t 2 + 40t + 5. How long does it take for the ball to reach its maximum height?

FAQs on Roots of Quadratic Equations

What is quadratic equation.

A quadratic equation is a polynomial equation of the second degree, typically written in the form ax 2 + bx + c = 0 , where a, b, and c are constants, and x is the variable.

What are the Roots of a Quadratic Equation?

The roots of a quadratic equation are the values of ‘x’ that satisfy the equation, making it equal to zero. These are the values where the graph of the quadratic function intersects the x-axis.

How do you Find the Roots of Quadratic Equation?

You can find the roots of a quadratic equation using the quadratic formula, i.e., x = (-b ± √(b 2 – 4ac)) / (2a). Other than this, we can also find roots using the methods such as factoring and completing the square.

How many Roots Can a Quadratic Equation Have?

A quadratic equation can have either two real roots, one real root (a repeated root), or two complex roots (conjugate pair of complex numbers).

Can a Quadratic Equation have no Real Roots?

Yes, if the discriminant is negative (b 2 – 4ac < 0), the quadratic equation will have two complex roots and no real roots.

How do you Determine the Nature of the Roots of Quadratic Equation?

You can determine the nature of the roots by calculating the discriminant (b 2 – 4ac) and examining its value in relation to zero. If discriminant > 0, two real roots, If discriminant = 0, one real repeated root, and If discriminant < 0, two complex roots i.e., no real roots.

What is the Sum and Product of the Roots of Quadratic Equation?

For a quadratic equation in the form ax 2 + bx + c = 0, the sum of the roots is -b/a, and the product of the roots is c/a.

What are Four Types of Roots of Quadratic Equations?

Four types of roots in quadratic equations are: Two Real and Distinct Roots Two Real and Equal Roots One Real and Repeated Root Two Complex Roots

Why are Alpha (α) and Beta (β) used to Represent Roots of Quadratic Equation?

Alpha (α) and beta (β) are commonly used to represent the roots of a quadratic equation for simplicity and consistency in mathematical notation.

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Roots for Quadratic Equation :=3x² + 12x + 36 Root 1 =-2 + 2.8284i Root 2 =-2 – 2.8284i —-Run complete. 15 symbols evaluated.—-

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The standard form of a quadratic equation is:

The solutions of this quadratic equation is given by:

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We have imported the cmath module to perform complex square root. First, we calculate the discriminant and then find the two solutions of the quadratic equation.

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