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solving quadratic equations problems

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Section 2.5 : Quadratic Equations - Part I

For problems 1 – 7 solve the quadratic equation by factoring.

For problems 8 & 9 use factoring to solve the equation.

For problems 10 – 12 use factoring to solve the equation.

For problems 13 – 16 use the Square Root Property to solve the equation.

solving quadratic equations problems

A quadratic equation is an equation that could be written as

ax 2 + bx + c = 0 

There are three basic methods for solving quadratic equations: factoring, using the quadratic formula, and completing the square. 

To solve a quadratic equation by factoring,

Solve x 2 – 6 x = 16. 

Following the steps, 

x 2 – 6 x = 16 becomes x 2 – 6 x – 16 = 0 

( x – 8)( x + 2) = 0 

solving quadratic equations problems

Both values, 8 and –2, are solutions to the original equation.

Solve y 2 = – 6 y – 5. 

Setting all terms equal to zero, 

y 2 + 6 y + 5 = 0 

( y + 5)( y + 1) = 0 

solving quadratic equations problems

To check, y 2 = –6 y – 5 

solving quadratic equations problems

A quadratic with a term missing is called an incomplete quadratic (as long as the ax 2 term isn't missing). 

Solve x 2 – 16 = 0. 

solving quadratic equations problems

To check, x 2 – 16 = 0 

solving quadratic equations problems

Solve x 2 + 6 x = 0. 

solving quadratic equations problems

To check, x 2 + 6 x = 0 

solving quadratic equations problems

Solve 2 x 2 + 2 x – 1 = x 2 + 6 x – 5. 

First, simplify by putting all terms on one side and combining like terms.

solving quadratic equations problems

Now, factor.

solving quadratic equations problems

To check, 2 x 2 + 2 x – 1 = x 2 + 6 x – 5 

solving quadratic equations problems

The quadratic formula

solving quadratic equations problems

a, b, and c are taken from the quadratic equation written in its general form of 

where a is the numeral that goes in front of x 2 , b is the numeral that goes in front of x , and c is the numeral with no variable next to it (a.k.a., “the constant”). 

When using the quadratic formula, you should be aware of three possibilities. These three possibilities are distinguished by a part of the formula called the discriminant. The discriminant is the value under the radical sign, b 2 – 4 ac . A quadratic equation with real numbers as coefficients can have the following:

Solve for x : x 2 – 5 x = –6. 

Setting all terms equal to 0, 

x 2 – 5 x + 6 = 0 

Then substitute 1 (which is understood to be in front of the x 2 ), –5, and 6 for a , b , and c, respectively, in the quadratic formula and simplify. 

solving quadratic equations problems

Because the discriminant b 2 – 4 ac is positive, you get two different real roots. 

Example produces rational roots. In Example , the quadratic formula is used to solve an equation whose roots are not rational. 

Solve for y : y 2 = –2y + 2. 

y 2 + 2 y – 2 = 0 

Then substitute 1, 2, and –2 for a , b , and c, respectively, in the quadratic formula and simplify. 

solving quadratic equations problems

Note that the two roots are irrational.

Solve for x : x 2 + 2 x + 1 = 0. 

Substituting in the quadratic formula,

solving quadratic equations problems

Since the discriminant b 2 – 4 ac is 0, the equation has one root. 

The quadratic formula can also be used to solve quadratic equations whose roots are imaginary numbers, that is, they have no solution in the real number system.

Solve for x : x ( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0. 

solving quadratic equations problems

Since the discriminant b 2 – 4 ac is negative, this equation has no solution in the real number system. 

solving quadratic equations problems

Completing the square

A third method of solving quadratic equations that works with both real and imaginary roots is called completing the square.

solving quadratic equations problems

Solve for x : x 2 – 6 x + 5 = 0. 

Arrange in the form of

solving quadratic equations problems

Take the square root of both sides.

x – 3 = ±2 

solving quadratic equations problems

Solve for y : y 2 + 2 y – 4 = 0. 

solving quadratic equations problems

Solve for x : 2 x 2 + 3 x + 2 = 0. 

solving quadratic equations problems

There is no solution in the real number system. It may interest you to know that the completing the square process for solving quadratic equations was used on the equation ax 2 + bx + c = 0 to derive the quadratic formula. 

Previous Quiz: Operations with Square Roots

Next Quiz: Solving Quadratic Equations

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Solving Quadratic Equations

Many word problems Involving unknown quantities can be translated for solving quadratic equations

Methods of solving quadratic equations are discussed here in the following steps.

Step I: Denote the unknown quantities by x, y etc.

Step II: use the conditions of the problem to establish in unknown quantities.

Step III: Use the equations to establish one quadratic equation in one unknown.

Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.

Now we will learn how to frame the equations from word problem:

1.  The product of two consecutive integers is 132. Frame an equation for the statement. What is the degree of the equation?

Method I: Using only one unknown

Let the two consecutive integers be x and x + 1

Form the equation, the product of x and x + 1 is 132.

Therefore, x(x + 1) = 132

⟹ x\(^{2}\) + x - 132 = 0, which is quadratic in x.

This is the equation of the statement, x denoting the smaller integer.

Method II: Using more than one unknown

Let the consecutive integers be x and y, x being the smaller integer.

As consecutive integers differ by 1, y - x = 1 ........................................... (i)

Again, from the question, the product of x and y is 132.

So, xy = 132 ........................................... (ii)

From (i), y = 1 + x.

Putting y = 1 + x in (ii),

x(1 + x) = 132

Solving the quadratic equation, we get the value of x. Then the value of y can be determined by substituting the value of x in y = 1 + x.

2. The length of a rectangle is greater than its breadth by 3m. If its area be 10 sq. m, find the perimeter.

Suppose, the breadth of the rectangle = x m.

Therefore, length of the rectangle = (x + 3) m.

So, area = (x + 3)x sq. m

Hence, by the condition of the problem

(x + 3)x = 10

⟹ x\(^{2}\) + 3x - 10 = 0

⟹ (x + 5)(x - 2) = 0

So, x = -5,2

But x = - 5 is not acceptable, since breadth cannot be negative.

Therefore x = 2

Hence, breadth = 2 m

and length = 5 m

Therefore, Perimeter = 2(2 + 5) m = 14 m.

x = -5 does not satisfy the conditions of the problem length or breadth can never be negative. Such a root is called an extraneous root. In solving a problem, each root of the quadratic equation is to be verified whether it satisfies the conditions of the given problem. An extraneous root is to be rejected.

Quadratic Equation

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Roots of a Quadratic Equation

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Quadratic Equations 

Word Problems on Quadratic Equations by Factoring

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring

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Quadratic Equations

An example of a Quadratic Equation :

The function makes 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:

Here are some examples:

Have a Play With It

Play with the " Quadratic Equation Explorer " so you can see:

Hidden Quadratic Equations!

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

ax 2 + bx + c = 0

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!

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:

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

And we see them on this graph.

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

√(−16) = 4 i (where i is the imaginary number √−1)

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

√(−9) = 3 i (where i is the imaginary number √−1)

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 Questions

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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:

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:

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:

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.

solving quadratic equations problems

Practice Questions

Solve these quadratic equations and find the roots. 

Frequently Asked Questions on Quadratic Equations

What is a quadratic equation, what are the examples of quadratic equations, what is the formula for quadratics, what are the methods to solve the quadratic equation, what are the roots of the quadratic equation, what are the zeroes of the quadratic equation.

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  6. (3.1 Extension) Applications of Quadratics

COMMENTS

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    Example: Solve 5x2 + 6x + 1 = 0 ; Quadratic Formula: · −b ± √(b2 − 4ac) 2a ; Put in a, b and c: · −6 ± √(62 − 4×5×1) 2×5 ; Solve: · −6 ± √(36− 20) 10 ; −6 ±

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