quadratic equation problems with solution and answer

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Example: 2x^2=18

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Quadratic Equation Worksheets (pdfs)

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Enjoy these free sheets. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key.

• Solve Quadratic Equations by Factoring
• Solve Quadratic Equations by Completing the Square
• Quadratic Formula Worksheet (real solutions)
• Quadratic Formula Worksheet (complex solutions)
• Quadratic Formula Worksheet (both real and complex solutions)
• Discriminant Worksheet
• Sum and Product of Roots
• Radical Equations Worksheet

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• ax^2+bx+c=0
• x^2+2x+1=3x-10
• 2x^2+4x-6=0
• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

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• High School Math Solutions – Quadratic Equations Calculator, Part 2 Solving quadratics by factorizing (link to previous post) usually works just fine. But what if the quadratic equation... Read More

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Quadratic Equation Questions

Quadratic equation questions are provided here for Class 10 students. A quadratic equation is a second-degree polynomial which is represented as ax 2 + bx + c = 0, where a is not equal to 0. Here, a, b and c are constants, also called coefficients and x is an unknown variable. Also, learn Quadratic Formula here.

Solving the problems based on quadratics will help students to understand the concept very well and also to score good marks in this section. All the questions are solved here step by step with a detailed explanation. In this article, we will give the definition and important formula for solving problems based on quadratic equations. The questions given here is in reference to the CBSE syllabus and NCERT curriculum. 

Definition of Quadratic Equation

Usually, the quadratic equation is represented in the form of ax 2 +bx+c=0, where x is the variable and a,b,c are the real numbers & a ≠ 0. Here, a and b are the coefficients of x 2 and x, respectively. So, basically, a quadratic equation is a polynomial whose highest degree is 2. Let us see some examples:

• 3x 2 +x+1, where a=3, b=1, c=1
• 9x 2 -11x+5, where a=9, b=-11, c=5

Roots of Quadratic Equations:

If we solve any quadratic equation, then the value we obtain are called the roots of the equation. Since the degree of the quadratic equation is two, therefore we get here two solutions and hence two roots.

There are different methods to find the roots of quadratic equation, such as:

• Factorisation
• Completing the square
• Using quadratic formula

Learn: Factorization of Quadratic equations

Quadratic Equation Formula:

The quadratic formula to find the roots of the quadratic equation is given by:

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

Where b 2 -4ac is called the discriminant of the equation.

Based on the discriminant value, there are three possible conditions, which defines the nature of roots as follows:

• two distinct real roots, if b 2 – 4ac > 0
• two equal real roots, if b 2 – 4ac = 0
• no real roots, if b 2 – 4ac < 0

Also, learn quadratic equations for class 10 here.

Quadratic Equations Problems and Solutions

1. Rahul and Rohan have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they both have now is 124. How to find out how many marbles they had to start with.

Solution: Say, the number of marbles Rahul had be x.

Then the number of marbles Rohan had = 45 – x.

The number of marbles left with Rahul after losing 5 marbles = x – 5

The number of marbles left with Rohan after losing 5 marbles = 45 – x – 5 = 40 – x

The product of number of marbles = 124

(x – 5) (40 – x) = 124

40x – x 2 – 200 + 5x = 124

– x 2 + 45x – 200 = 124

x 2 – 45x + 324 = 0

This represents the quadratic equation. Hence by solving the given equation for x, we get;

x = 36 and x = 9

So, the number of marbles Rahul had is 36 and Rohan had is 9 or vice versa.

2. Check if x(x + 1) + 8 = (x + 2) (x – 2) is in the form of quadratic equation.

Solution: Given,

x(x + 1) + 8 = (x + 2) (x – 2)

Cancel x 2 both the sides.

Since, this expression is not in the form of ax 2 +bx+c, hence it is not a quadratic equation.

3. Find the roots of the equation 2x 2 – 5x + 3 = 0 using factorisation.

2x 2 – 5x + 3 = 0

2x 2 – 2x-3x+3 = 0

2x(x-1)-3(x-1) = 0

(2x-3) (x-1) = 0

2x-3 = 0; x = 3/2

(x-1) = 0; x=1

Therefore, 3/2 and 1 are the roots of the given equation.

4. Solve the quadratic equation 2x 2 + x – 300 = 0 using factorisation.

Solution: 2x 2 + x – 300 = 0

2x 2 – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

(x – 12)(2x + 25) = 0

x-12=0; x=12

(2x+25) = 0; x=-25/2 = -12.5

Therefore, 12 and -12.5 are two roots of the given equation.

Also, read   Factorisation .

5. Solve the equation x 2 +4x-5=0.

x 2 + 4x – 5 = 0

x 2 -1x+5x-5 = 0

x(x-1)+5(x-1) =0

(x-1)(x+5) =0

Hence, (x-1) =0, and (x+5) =0

similarly, x+5 = 0

x=-5 & x=1

6. Solve the quadratic equation 2x 2 + x – 528 = 0, using quadratic formula.

Solution: If we compare it with standard equation, ax 2 +bx+c = 0

a=2, b=1 and c=-528

Hence, by using the quadratic formula:

Now putting the values of a,b and c.

x=64/4 or x=-66/4

x=16 or x=-33/2

7. Find the roots of x 2 + 4x + 5 = 0, if any exist, using quadratic formula.

Solution: To check whether there are real roots available for the quadratic equation, we need the find the discriminant value.

D = b 2 -4ac = 4 2 – 4(1)(5) = 16-20 = -4

Since the square root of -4 will not give a real number. Hence there is no real roots for the given equation.

8. Find the discriminant of the equation: 3x 2 -2x+⅓ = 0.

Solution: Here, a = 3, b=-2 and c=⅓

Hence, discriminant, D = b 2 – 4ac

D = (-2) 2 -4(3)(⅓)

Video Lesson

Quadratic equation worksheet.

Practice Questions

Solve these quadratic equations and find the roots. 

• x 2 -5x-14=0         [Answer: x=-2 & x=7]
• X 2  = 11x -28       [Answer: x=4 & x = 7]
• 6x 2 – x = 5            [Answer: x=-⅚ & x = 1]
• 12x 2 = 25x          [Answer: x=0 & x=25/12]

Frequently Asked Questions on Quadratic Equations

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Real World Examples of Quadratic Equations

A Quadratic Equation looks like this:

Quadratic equations pop up in many real world situations!

Here we have collected some examples for you, and solve each using different methods:

• Factoring Quadratics
• Completing the Square
• Graphing Quadratic Equations
• The Quadratic Formula
• Online Quadratic Equation Solver

Each example follows three general stages:

• Take the real world description and make some equations
• Use your common sense to interpret the results

Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

... and a Quadratic Equation tells you its position at all times!

Example: Throwing a Ball

A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. when does it hit the ground.

Ignoring air resistance, we can work out its height by adding up these three things: (Note: t is time in seconds)

Add them up and the height h at any time t is:

h = 3 + 14t − 5t 2

And the ball will hit the ground when the height is zero:

3 + 14t − 5t 2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t 2 + 14t + 3 = 0

It looks even better when we multiply all terms by −1 :

5t 2 − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give a×c , and add to give b " method in Factoring Quadratics :

a×c = − 15 , and b = − 14 .

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

The "t = −0.2" is a negative time, impossible in our case.

The "t = 3" is the answer we want:

The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = −5t 2 + 14t + 3

It shows you the height of the ball vs time

Some interesting points:

(0,3) When t=0 (at the start) the ball is at 3 m

(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.

(3,0) says that at 3 seconds the ball is at ground level.

Also notice that the ball goes nearly 13 meters high.

Note: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations , and has two steps:

Find where (along the horizontal axis) the top occurs using −b/2a :

• t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds

Then find the height using that value (1.4)

• h = −5t 2 + 14t + 3 = −5(1.4) 2 + 14 × 1.4 + 3 = 12.8 meters

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

Your costs are going to be:

• $700,000 for manufacturing set-up costs, advertising, etc
• $110 to make each bike

Based on similar bikes, you can expect sales to follow this "Demand Curve":

• Unit Sales = 70,000 − 200P

Where "P" is the price.

For example, if you set the price:

• at $0, you just give away 70,000 bikes
• at $350, you won't sell any bikes at all
• at $300 you might sell 70,000 − 200×300 = 10,000 bikes

So ... what is the best price? And how many should you make?

Let us make some equations!

How many you sell depends on price, so use "P" for Price as the variable

• Sales in Dollars = Units × Price = (70,000 − 200P) × P = 70,000P − 200P 2
• Costs = 700,000 + 110 x (70,000 − 200P) = 700,000 + 7,700,000 − 22,000P = 8,400,000 − 22,000P
• Profit = Sales-Costs = 70,000P − 200P 2 − (8,400,000 − 22,000P) = −200P 2 + 92,000P − 8,400,000

Profit = −200P 2 + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square .

Solve: −200P 2 + 92,000P − 8,400,000 = 0

Step 1 Divide all terms by -200

Step 2 Move the number term to the right side of the equation:

Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2) 2 = (−460/2) 2 = (−230) 2 = 52900

Step 4 Take the square root on both sides of the equation:

Step 5 Subtract (-230) from both sides (in other words, add 230):

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don't we?

It is exactly half way in-between! At $230

And here is the graph:

The best sale price is $230 , and you can expect:

• Unit Sales = 70,000 − 200 x 230 = 24,000
• Sales in Dollars = $230 x 24,000 = $5,520,000
• Costs = 700,000 + $110 x 24,000 = $3,340,000
• Profit = $5,520,000 − $3,340,000 = $2,180,000

A very profitable venture.

Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm 2

The inside of the frame has to be 11 cm by 6 cm

What should the width x of the metal be?

Area of steel before cutting:

Area of steel after cutting out the 11 × 6 middle:

Let us solve this one graphically !

Here is the graph of 4x 2 + 34x :

The desired area of 28 is shown as a horizontal line.

The area equals 28 cm 2 when:

x is about −9.3 or 0.8

The negative value of x make no sense, so the answer is:

x = 0.8 cm (approx.)

Example: River Cruise

A 3 hour river cruise goes 15 km upstream and then back again. the river has a current of 2 km an hour. what is the boat's speed and how long was the upstream journey.

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:

• Let x = the boat's speed in the water (km/h)
• Let v = the speed relative to the land (km/h)

Because the river flows downstream at 2 km/h:

• when going upstream, v = x−2 (its speed is reduced by 2 km/h)
• when going downstream, v = x+2 (its speed is increased by 2 km/h)

We can turn those speeds into times using:

time = distance / speed

(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)

And we know the total time is 3 hours:

total time = time upstream + time downstream = 3 hours

Put all that together:

total time = 15/(x−2) + 15/(x+2) = 3 hours

Now we use our algebra skills to solve for "x".

First, get rid of the fractions by multiplying through by (x-2) (x+2) :

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x 2 −4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x 2 − 30x − 12 = 0

It is a Quadratic Equation! Let us solve it using the Quadratic Formula :

Where a , b and c are from the Quadratic Equation in "Standard Form": ax 2 + bx + c = 0

Solve 3x 2 - 30x - 12 = 0

Answer: x = −0.39 or 10.39 (to 2 decimal places)

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

Example: Resistors In Parallel

Two resistors are in parallel, like in this diagram:

The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.

What are the values of the two resistors?

The formula to work out total resistance "R T " is:

1 R T   =   1 R 1 + 1 R 2

In this case, we have R T = 2 and R 2 = R 1 + 3

1 2   =   1 R 1 + 1 R 1 +3

To get rid of the fractions we can multiply all terms by 2R 1 (R 1 + 3) and then simplify:

Yes! A Quadratic Equation !

Let us solve it using our Quadratic Equation Solver .

• Enter 1, −1 and −6
• And you should get the answers −2 and 3

R 1 cannot be negative, so R 1 = 3 Ohms is the answer.

The two resistors are 3 ohms and 6 ohms.

Quadratic Equations are useful in many other areas:

For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.

Quadratic equations are also needed when studying lenses and curved mirrors.

And many questions involving time, distance and speed need quadratic equations.

Quadratic Formula Calculator

Enter the equation you want to solve using the quadratic formula.

The Quadratic Formula Calculator finds solutions to quadratic equations with real coefficients. For equations with real solutions, you can use the graphing tool to visualize the solutions.

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

Click the blue arrow to submit. Choose "Solve Using the Quadratic Formula" from the topic selector and click to see the result in our Algebra Calculator !

Solve Using the Quadratic Formula Apply the Quadratic Formula

Popular Problems

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How to Solve Quadratic Equations

Last Updated: February 10, 2023 Fact Checked

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 1,365,198 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.

Factoring the Equation

• 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.

Using the Quadratic Formula

• 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)

Completing the Square

• 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)

Practice Problems and Answers

Expert Q&A

• 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 1 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
• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Thanks Helpful 2 Not Helpful 0

You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
• ↑ http://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
• ↑ http://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 Questions

Basic Quadratic Equation Questions

Quadratic equations are an important part of algebra, and as students, we must all be familiar with their definition and the ways of solving quadratic equation problems. In this article, we are going to familiarize the students with all the concepts surrounding quadratic equations and the methods of solving problems related to this topic. A short definition of a quadratic equation would be: a quadratic equation is a second-degree polynomial, which we represent as ‘ax 2 + bx + c’ in general.

In this representation, a cannot be equal to 0 and b,c are known as coefficients and are constant by nature. With this basic introduction, let's move forward with a formal definition, formulae and detailed solutions to quadratic equation questions to enable better understanding.

Definition of Quadratic Equations

A quadratic equation is a polynomial where the highest power of the variable is 2. We generally represent it as ax 2 + bx + c. Here a, b and c are real numbers or constants, and x is the variable. In this case, the value of a cannot be 0 as that would remove the x 2 term, and the equation won't be quadratic after that.

A quadratic equation is an equation of second degree with more than two terms. It means that at least one of the terms of the equation is squared. In the above-given equation. The answer to the equation also known as the roots of the equation is the value of the “x”. The value of the “x” has to satisfy the equation. 

Some examples of quadratic equations can be as follows:

56x 2 + ⅔ x + 1, where a = 56, b = ⅔ and c = 1.

-4/3 x 2 + 64x - 30, where a = -4/3, b = 64 and c = -30.

Roots of a Quadratic Equation

To solve basic quadratic equation questions or any quadratic equation problems, we need to solve the equation. Solving quadratic equations gives us the roots of the polynomial. The roots of the equation are the values of x at which ax 2 + bx + c = 0. Since a quadratic equation is a polynomial of degree 2, we obtain two roots in this case.

There are several methods for solving quadratic equation problems, as we can see below:

Factorization Method.

Completing The Square Method.

Quadratic Equation Formula.

Quadratic Equation Formula

So what is the quadratic equation formula? The quadratic equation formula or the Sridharacharya Formula is a method for finding out the roots of two-degree polynomials. This formula helps solve quadratic equation problems. The formula is as given below:

\[ x = \frac{-b \pm\sqrt{b^{2}-4ac}}{2a} \]

Where x represents the roots of the equation and (b 2 −4ac) is the discriminant.

By finding out the value of the discriminant, we can predict the nature of the roots. There are three possibilities with three different implications:

Two distinct roots which are real, if b 2 - 4ac > 0.

Two real roots equal in magnitude, if b 2 - 4ac = 0.

Imaginary roots or absence of real roots if b 2 - 4ac < 0.

Now that the basic principles of quadratic equations are clear, we will move on to some solved examples. But before that, let us list some quadratic equation questions for the students to solve.

Quadratic Equation Practice Questions

The following are a list of questions for you to solve once you have gone through the quadratic equation questions and answers in the solved examples section:

Find the determinant of the following quadratic equations: 2x 2 + 3x + 6, 70x 2 + 49 + 14, ⅔ y 2 + 63y + 42.

Find the roots of the following quadratic equations: x 2 - 45x + 324, 2x 2 - 22x + 42, ½ x 2 + 2x + 4.

The product of two consecutive numbers is 420, and their sum is 41. Find the numbers.

Before solving these, let's check out the solved examples with questions and answers on the quadratic equation.

Solved Examples

Q1. Solve x 2 + 5x + 6 = 20 by factorization method.

Answer: The given polynomial or quadratic equation is

x 2 + 5x + 6 = 20

Solving by factorization method, 

x 2 + 5x + 6 - 20 = 0.

or, x 2 + 5x - 14 = 0

or, x 2 - 2x + 7x - 14 = 0

or, x(x - 2) +7(x - 2) = 0

or, (x - 2)(x + 7) = 0

or, (x - 2) = 0, (x + 7) = 0 

or, x = +2, -7.

Q2. Solve 2x 2 - 5x + 3 using the quadratic equation formula.

Answer: The quadratic equation formula is:

\[ x = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \]

The determinant or b 2 -4ac = (-5) 2 - 4 × 3 × 2 = 25 - 24 = 1

\[ \sqrt{b^{2} - 4ac} = 1 \]

Therefore, \[ x= \frac{-(-5)\pm1}{2\times2} \]

\[ x = \frac{5+1}{4} = \frac{6}{4} = \frac{3}{2} \]

\[ x = \frac{5-1}{4} = \frac{4}{4} = 1 \]

Thus the roots of the equation are 3/2 and 1.

This is all about the roots of quadratic equations and their formulas. Learn the formulas and find out how they are used to derive the roots of an equation easily. 

FAQs on Quadratic Equation Questions

1. Define a quadratic equation along with suitable examples. also, state the quadratic equation formula.

A quadratic equation is a polynomial where the highest power of the variable is neither more nor less than 2. So essentially, a quadratic equation is a polynomial of degree 2. We represent such an equation in a general format as ax 2 + bx + c, where a, b and c are known as the coefficients or the constants of the equation. The thumb rule for quadratic equations is that the value of a cannot be 0. The x in the expression is the variable. This algebraic expression, when solved, will yield two roots.

Some examples of quadratic equations are:

3x 2 + 4x + 7 = 34

x 2 + 8x + 12 = 40

The quadratic equation formula is a method for solving quadratic equation questions. The formula is as follows:

where x represents the roots of the equation.

2. What are the roots of a quadratic equation? What are the zeroes of a polynomial?  

The roots of a quadratic equation are the values obtained when we solve the equation. They are those values of x for which the expression ax 2 +bx+c becomes equal to 0. These values are also known as the zeroes of the polynomial. Since a quadratic equation is essentially a polynomial of degree 2, we get two roots after solving the given polynomial.

As we practice more and more quadratic equation sums, our ideas regarding which method to use while solving a given question will get clearer.

3. How can I solve a quadratic equation?

Quadratic equation is an equation with more than one term in it and at least one of the terms having degree 2. Its general form is ax 2  + bx + c, whereas a,b,c are real numbers and a is not equal to zero. The values which satisfy the “x” in the equation are the solution for the quadratic equation. They are also known as the roots of the equation. The quadratic equation can be solved in the algebraic method and graphical method.

In the algebraic method, the equation is reduced to the roots by shifting terms from L.H.S to the R.H.S and using different mathematical operations. In the graphic method, the equation is solved by drawing it on the map and solving it using the parabola the equation makes on the graph. The value of “a” determines whether the graph of the equation is concave parabola or convex parabola. The value of the discriminant decides whether the curve will intersect the x-axis or not.

4.  What are the applications of quadratic equations?

A quadratic equation is a polynomial equation with more than one term and at least one of the terms having 2 as square. It is generally used in different situations in day-to-day life. In constructing rooms and boxes of different geometric shapes. If you want to construct a box made of wood with 5 square feet dimensions, you can write a quadratic equation to measure its area and calculate the material required. 

It can also be used in selling something and calculating the profit and loss you may incur after selling the good. To know it, you can simply form a quadratic equation. It can also be used in athletics while throwing objects like a javelin, shot put ball, etc. It can also be used to calculate the distance. Generally, when someone travels up and down the river uses this equation to measure the distance to be travelled. 

5. How are quadratic equations used in athletics and construction?

Quadratic equations are equations with at least one term having 2 as a square and it has more than one term in the equation preferably, four terms. The general form of the equation is ax 2  + bx+c whereas a,b,c are real numbers and a is not equal to zero. The values which satisfy the “x” in the equation are the solution for the quadratic equation. They are also known as the roots of the equation. In exams, they have preferable weightage and practising them correctly will improve the marks. On the other hand, they are also used in real-life situations. In athletics, it is used to measure the speed and force to be applied to throw an object like an arrow, shot put the ball, discus, etc. They use the velocity equation to measure the height of the ball from which it should be thrown. In the field of construction, a quadratic equation is framed with the known dimensions of the building or a room to figure out unknown values like the area, perimeter, etc.