College Algebra
Tutorial 17: Quadratic Equations

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Learning Objectives

After completing this tutorial, you should be able to:

Solve quadratic equations by factoring.
Solve quadratic equations by the square root method.
Solve quadratic equations by completing the square.
Solve quadratic equations by using the quadratic formula.
Find the discriminant of a quadratic equation and use it to tell how many and what type of solutions the equation has.

Introduction

In this tutorial, we will be looking at solving a specific type of equation called the quadratic equation. The methods of solving these types of equations that we will take a look at are solving by factoring, by using the square root method, by completing the square, and by using the quadratic equation. Sometimes one method won't work or another is just faster, depending on the quadratic equation given. So it is good to know different ways to solve quadratic equations so you will be prepared for any type of situation. After completing this tutorial, you will be a master at solving quadratic equations. Solving equations in general is a very essential part of Algebra. So I guess we better get to it.

Tutorial

Quadratic Equation
Standard form:

,
Where a does not equal 0.

Note that in Tutorial 14: Linear Equations in One Variable, we learned that a linear equation can be written in the form ax + b = 0 and that the exponent on the variable was 1.

Note that the difference is the highest exponent on the variable on the quadratic equation is 2.

We are going to talk about four ways to solve quadratics.

Solving Quadratic Equations by Factoring

You can solve a quadratic equation by factoring if, after writing it in standard form, the quadratic expression factors.

Step 1: Simplify each side if needed.

This would involve things like removing ( ), removing fractions, adding like terms, etc.

To remove ( ): Just use the distributive property.

To remove fractions: Since fractions are another way to write division, and the inverse of divide is to multiply, you remove fractions by multiplying both sides by the LCD of all of your fractions.

Step 2: Write in standard form, , if needed.

If it is not in standard form, move any term(s) to the appropriate side by using the addition/subtraction property of equality.

Also, make sure that the squared term is written first left to right, the x term is second and the constant is third and it is set equal to 0.

Step 3: Factor.

If you need a review on factoring go to Tutorial 7: Factoring Polynomials.

Step 4: Use the Zero-Product Principle.

If ab = 0, then a = 0 or b = 0.

0 is our magic number because the only way a product can become 0 is if at least one of its factors is 0.

You can not guarantee what the factors would have to be if the product was set equal to any other number. For example if ab = 1, then a = 5 and b = 1/5 or a = 3 and b = 1/3, etc. But with the product set equal to 0, we can guarantee finding the solution by setting each factor equal to 0. That is why it is important to get it in standard form to begin with.

Step 5: Solve for the linear equation(s) set up in step 4.

If a quadratic equation factors, it will factor into either one linear factor squared or two distinct linear factors. So, the equations found in step 4 will be linear equations. If you need a review on solving linear equations feel free to go to Tutorial 14: Linear Equations in One Variable.

Example 1: Solve by factoring.

Step 1: Simplify each side if needed.

This quadratic equation is already simplified.

Step 2: Write in standard form, , if needed.

This quadratic equation is already in standard form.

Step 3: Factor.

*Quad. eq. in standard form
*Factor the trinomial

Step 4: Use the Zero-Product Principle

AND

Step 5: Solve for the linear equation(s) set up in step 4.

*Use Zero-Product Principle
*Solve the first linear equation

*Solve the second linear equation

There are two solutions to this quadratic equation: x = -5 and x = 2.

Example 2: Solve by factoring.

Step 1: Simplify each side if needed.

*Mult. both sides by LCD of 6 to clear fractions

Step 2: Write in standard form, , if needed.

*Inverse of add. 16 is sub. 16
*Quad. eq. in standard form

Step 3: Factor.

*Quad. eq. in standard form
*Factor the diff. of two squares

Step 4: Use the Zero-Product Principle

AND

Step 5: Solve for the linear equation(s) set up in step 4.

*Use Zero-Product Principle
*Solve the first linear equation

*Solve the second linear equation

There are two solutions to this quadratic equation: x = -4/5 and x = 4/5.

Example 3: Solve by factoring.

Step 1: Simplify each side if needed.

*Use Dist. Prop. to clear the ( )

Step 2: Write in standard form, , if needed.

*Inverse of add. 2 is sub. 2
*Quad. eq. in standard form

Step 3: Factor.

*Quad. eq. in standard form
*Factor the trinomial

Step 4: Use the Zero-Product Principle

AND

Step 5: Solve for the linear equation(s) set up in step 4.

*Use Zero-Product Principle
*Solve the first linear equation

*Solve the second linear equation

There are two solutions to this quadratic equation: x = -2/3 and x = 1/2.

Solving Quadratic Equations by the Square Root Method

You can solve a quadratic equation by the square root method if you can write it in the form .

Step 1: Write the quadratic equation in the form if needed.

A and B represent algebraic expressions. When you have the quadratic equation written in this form, it allows you to use the square root method described in step 2.

If it is not in this form, , move any term(s) to the appropriate side by using the addition/subtraction or multiplication/division property of equality.

Step 2: Apply the square root method.

If A and B are algebraic expressions such that

, then

also written .

In other words, if you have an expression squared set equal to another expression, the inverse operation to solve it is to take the square root of both sides. Since both a positive and its opposite squared result in the same answer, then you will have two answers, plus or minus the square root of B.

Step 3: Solve for the linear equation(s) set up in step 2.

After applying the square root method to a quadratic equation you will end up with either one or two linear equations to solve. Most times you will have two linear equations, but if B is equal to 0, then you will only have one since plus or minus 0 is only one number. If you need a review on solving linear equations feel free to go to Tutorial 14: Linear Equations in One Variable.

Example 4: Solve by using the square root method.

Step 1: Write the quadratic equation in the form if needed

AND

Step 2: Apply the square root method.

*Written in the form

*Apply the sq. root method
*There are 2 solutions

Step 3: Solve for the linear equation(s) set up in step 2.

*Sq. root of 16 = 4

*Neg. sq. root of 16 = - 4

There are two solutions to this quadratic equation: x = 4 and x = -4.

Example 5: Solve by using the square root method.

Step 1: Write the quadratic equation in the form if needed

AND

Step 2: Apply the square root method.

Note how this quadratic equation is not in the form

to begin with. The 5 is NOT part of the expression being squared on the left side of the equation. We can easily write it in the form

by dividing both sides by 5.

*Not in the form

*Inv. of mult. by 5 is div. by 5

*Written in the form

*Apply the sq. root method
*There are 2 solutions

Step 3: Solve for the linear equation(s) set up in step 2.

*Sq. root of 4 = 2

*Neg. sq. root of 4 = -2

There are two solutions to this quadratic equation: x = 2 and x = -2.

Example 6: Solve by using the square root method.

Step 1: Write the quadratic equation in the form if needed

AND

Step 2: Apply the square root method.

*Written in the form

*Apply the sq. root method
*There are 2 solutions

Step 3: Solve for the linear equation(s) set up in step 2.

*Sq. root of 20 = 2 sq. root of 5
*Solve for x

*Neg. sq. root of 20 = -2 sq. root of 5
*Solve for x

There are two solutions to this quadratic equation: x = and x = .

Solving Quadratic Equations by Completing the Square

You can solve ANY quadratic equation by completing the square. This comes in handy when a quadratic equation does not factor or is difficult to factor.

Step 1: Make sure that the coefficient on the term is equal to 1.

If the coefficient of the

term is already 1, then proceed to step 2.

If the coefficient of the term is not equal to 1, then divide both sides by that coefficient.

Step 2: Isolate the and x terms.

In other words, rewrite it so that the

and x terms are on one side and the constant is on the other side.

Step 3: Complete the square.

At this point we will be creating a perfect square trinomial (PST). Recall that a PST is a trinomial of the form

and it factors in the form

. When it is in that form it will allow us to continue onto the next step and take the square root of both sides and find a solution.

We need to find a number that we can add to the and x terms so that we have a PST.

We can get that magic number by doing the following:

If we have

we can complete it’s square by adding the constant

In other words, we complete the square by taking ½ of b (the coefficient of the x term) and then squaring it. Make sure you remember to add it to BOTH sides to keep the equation balanced.

Step 4: Factor the perfect square trinomial (created in step 3) as a binomial squared.

If you need a review on factoring a perfect square trinomial, feel free to go to Tutorial 7: Factoring Polynomials.

Step 5: Solve the equation in step 4 by using the square root method.

Example 7: Solve by completing the square.

Step 1:Make sure that the coefficient on the term is equal to 1.

The coefficient of the

term is already 1.

Step 2: Isolate the and x terms.

The

and x terms are already isolated.

Step 3: Complete the square.

*b is the coefficient of the x term

*Complete the square by taking 1/2 of b and squaring it

*Add constant found above to BOTH sides of the eq.

*This creates a PST on the left side of eq.

Step 4: Factor the perfect square trinomial (created in step 3) as a binomial squared.

*Factor the PST

Step 5: Solve the equation in step 4 by using the square root method.

*Written in the form

*Apply the sq. root method
*There are 2 solutions

There are two solutions to this quadratic equation: x = 9 and x = 1.

Example 8: Solve by completing the square.

Step 1:Make sure that the coefficient on the term is equal to 1.

Note how the coefficient on the

term is not 1 to begin with. We can easily fix that by dividing both sides by that coefficient, which in this case is 3 .

*Divide both sides by 3

*Coefficient of term is now 1

Step 2: Isolate the and x terms.

Note how the

and x terms are not isolated to begin with. We can easily fix that by moving the constant to the other side of the equation.

*Inverse of add. 3 is sub. 3

* and x terms are now isolated

Step 3: Complete the square.

*b is the coefficient of the x term

*Complete the square by taking 1/2 of b and squaring it

*Add constant found above to BOTH sides of the eq.

*This creates a PST on the left side of eq.

Step 4: Factor the perfect square trinomial (created in step 3) as a binomial squared.

*Factor the PST

Step 5: Solve the equation in step 4 by using the square root method.

*Written in the form

*Apply the sq. root method
*There are 2 solutions

*Square root of a negative 1 is i

There are two solutions to this quadratic equation: x = and x = .

Solving Quadratic Equations by the Quadratic Formula

When , then

You can solve ANY quadratic equation by using the quadratic formula. This comes in handy when a quadratic equation does not factor or is difficult to factor.

Step 1: Simplify each side if needed.

This would involve things like removing ( ), removing fractions, adding like terms, etc.

To remove ( ): Just use the distributive property.

Step 2: Write in standard form, , if needed.

If it is not in standard form, move any term(s) to the appropriate side by using the addition/subtraction property of equality.

Also, make sure that the squared term is written first left to right, the x term is second and the constant is third and it is set equal to 0.

Step 3: Identify a, b, and c.

When the quadratic equation is in standard form,

, then a is the coefficient in front of the

term, b is the coefficient in front of the x term, and c is the constant term.

Step 4: Plug the values found in step 3 into the quadratic formula.

When

, then

Step 5: Simplify if possible.

Example 9: Solve by using the quadratic formula.

Step 1: Simplify each side if needed.

This quadratic equation is already simplified.

Step 2: Write in standard form, , if needed.

This quadratic equation is already in standard form.

Step 3: Identify a, b, and c.

a, the number in front of x squared, is 2.

b, the number in front of x, is -5.

c, the constant, is 1.

Make sure that you keep the sign that is in front of each of these numbers.

Next we will plug it into the quadratic formula. Note that we are only plugging in numbers, we don't also plug in the variable.

Step 4: Plug the values found in step 3 into the quadratic formula

AND

Step 5: Simplify if possible.

*Quadratic formula

*Plug in values found above for a, b, and c

*Simplify

Example 10: Solve by using the quadratic formula.

Step 1: Simplify each side if needed.

This quadratic equation is already simplified.

Step 2: Write in standard form, , if needed.

This quadratic equation is already in standard form.

Step 3: Identify a, b, and c.

a, the number in front of x squared, is 1.

b, the number in front of x, is 0.
Note that b is 0 because the x term is missing.

c, the constant, is 9.

Make sure that you keep the sign that is in front of each of these numbers.

Next we will plug it into the quadratic formula. Note that we are only plugging in numbers, we don't also plug in the variable.

Step 4: Plug the values found in step 3 into the quadratic formula

AND

Step 5: Simplify if possible.

*Quadratic formula

*Plug in values found above for a, b, and c

*Simplify

*Square root of a negative 1 is i

Example 11: Solve by using the quadratic formula.

Step 1: Simplify each side if needed.

This quadratic equation is already simplified.

Step 2: Write in standard form, , if needed.

*Inverse of add. 6x is sub. 6x

*Quad. eq. in standard form

Step 3: Identify a, b, and c.

a, the number in front of x squared, is 1.

b, the number in front of x, is -6.

c, the constant, is 9.

Make sure that you keep the sign that is in front of each of these numbers.

Next we will plug it into the quadratic formula. Note that we are only plugging in numbers, we don't also plug in the variable.

Step 4: Plug the values found in step 3 into the quadratic formula

AND

Step 5: Simplify if possible.

*Quadratic formula

*Plug in values found above for a, b, and c

*Simplify

Discriminant

When a quadratic equation is in standard form, , the expression, , that is found under the square root part of the quadratic formula is called the discriminant.

The discriminant can tell you how many solutions there are going to be and if the solutions are real numbers or complex imaginary numbers.

Discriminant	Kinds of solution for
	Two distinct real solutions Note that the value of the discriminant is found under the square root and there is a + or - in front of it. So, if that value is positive, then there would be two distinct real number answers.
	One real solution Note that the value of the discriminant is found under the square root and there is a + or - in front of it. So, if that value is zero, + or - zero is the same number, so there would be only one real number solution.
	Two distinct complex imaginary solution Note that the value of the discriminant is found under the square root and there is a + or - in front of it. So, if that value is negative, then there would be two distinct complex imaginary number answers.

Finding the Discriminant

Step 1: Simplify each side if needed.

This would involve things like removing ( ), removing fractions, adding like terms, etc.

To remove ( ): Just use the distributive property.

Step 2: Write in standard form, , if needed.

If it is not in standard form, move any term(s) to the appropriate side by using the addition/subtraction property of equality.

Also, make sure that the squared term is written first left to right, the x term is second and the constant is third and it is set equal to 0.

Step 3: Identify a, b, and c.

When the quadratic equation is in standard form,

, then a is the coefficient in front of the

term, b is the coefficient in front of the x term, and c is the constant term.

Step 4: Plug the values found in step 3 into the discriminant, .

Step 5: Simplify if possible.

Example 12: Find the discriminant. Based on the discriminant, indicate how many and what type of solutions there would be.

Step 1: Simplify each side if needed.

This quadratic equation is already simplified.

Step 2: Write in standard form, , if needed.

This quadratic equation is already in standard form.

Step 3: Identify a, b, and c.

a, the number in front of x squared, is 3.

b, the number in front of x, is 1.

c, the constant, is 10.

Make sure that you keep the sign that is in front of each of these numbers.

Step 4: Plug the values found in step 3 into the discriminant, .

AND

Step 5: Simplify if possible.

*Discriminant formula

*Plug in values found above for a, b, and c

*Discriminant

Since the discriminant is a negative number, that means there are two distinct complex imaginary solutions.

Example 13: Find the discriminant. Based on the discriminant, indicate how many and what type of solutions there would be. .

Step 1: Simplify each side if needed.

This quadratic equation is already simplified.

Step 2: Write in standard form, , if needed.

*Inverse of sub. 16 is add. 16

*Quad. eq. in standard form

Step 3: Identify a, b, and c.

a, the number in front of x squared, is 1.

b, the number in front of x, is -8.

c, the constant, is 16.

Make sure that you keep the sign that is in front of each of these numbers.

Step 4: Plug the values found in step 3 into the discriminant, .

AND

Step 5: Simplify if possible.

*Discriminant formula

*Plug in values found above for a, b, and c

*Discriminant

Since the discriminant is zero, that means there is only one real number solution.

Example 14: Find the discriminant. Based on the discriminant, indicate how many and what type of solutions there would be. .

Step 1: Simplify each side if needed.

This quadratic equation is already simplified.

Step 2: Write in standard form, , if needed.

*Inverse of sub. 7x is add. 7x

*Quad. eq. in standard form

Step 3: Identify a, b, and c.

a, the number in front of x squared, is -5.

b, the number in front of x, is 7.

c, the constant, is 0.
Note that since the constant is missing it is understood to be 0.

Make sure that you keep the sign that is in front of each of these numbers.

Step 4: Plug the values found in step 3 into the discriminant, .

AND

Step 5: Simplify if possible.

*Discriminant formula

*Plug in values found above for a, b, and c

*Discriminant

Since the discriminant is a positive number, that means there are two distinct real number solutions.

Practice Problems

These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice.

To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.

1a.

(answer/discussion to 1a)

1b.

(answer/discussion to 1b)

2a.

(answer/discussion to 2a)

2b.

(answer/discussion to 2b)

3a.

(answer/discussion to 3a)

3b.

(answer/discussion to 3b)

4a.

(answer/discussion to 4a)

4b.

(answer/discussion to 4b)

4c.

(answer/discussion to 4c)

5a.

(answer/discussion to 5a)

5b.

(answer/discussion to 5b)

5c.

(answer/discussion to 5c)

Need Extra Help on these Topics?

The following are webpages that can assist you in the topics that were covered on this page:

http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut30_eqfact.htm
This webpage helps you with solving polynomial equations by factoring.

http://www.purplemath.com/modules/solvquad.htm
This webpage helps with solving quadratic equations by factoring

http://www.sosmath.com/algebra/quadraticeq/sobyfactor/ sobyfactor.html
This webpage helps with solving quadratic equations by factoring.

http://www.mathpower.com/tut99.htm
This webpage helps with solving quadratic equations by factoring.

http://www.mathpower.com/tut105.htm
This webpage helps with solving quadratic equations by factoring.

http://www.mathpower.com/tut110.htm
This webpage helps with solving quadratic equations by factoring.

http://www.purplemath.com/modules/solvquad2.htm
This webpage helps with solving quadratic equations by taking the square root.

http://www.purplemath.com/modules/solvquad3.htm
This webpage helps with solving quadratic equations by completing the square.

http://www.purplemath.com/modules/solvquad4.htm
This webpage helps with solving quadratic equations by using the quadratic formula.

Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.

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Last revised on Dec. 17, 2009 by Kim Seward.
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