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Algebra 1 /

Chapter 8

Solving Quadratic Equations Using Square Roots

Understanding quadratic equations is a foundational skill in algebra. This lesson introduces learners to a specific type of quadratic equation where the linear coefficient is zero, often referred to as simple quadratic equations. By isolating the x^2 term and applying square roots to both sides, one can find the solutions or roots of the equation. The lesson emphasizes the importance of considering both the principal and negative roots to ensure all potential solutions are identified. Real-world scenarios, such as calculating the area of a rectangle, are used to demonstrate the application of these equations, making the learning process engaging and relatable.

Lesson Settings & Tools

	11 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Image Credits

Slide of 11

This lesson will discuss how to solve a particular type of quadratic equation by using square roots.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Determining the Number of Solutions

Consider the following quadratic equations.

It is known that a quadratic equation can have none, one, or two solutions. Can the number of solutions of each of the above equations be determined without actually solving the equations?

Discussion

Simple Quadratic Equations

A quadratic equation is a polynomial equation of degree $2 .$ There is a special name for quadratic equations whose linear coefficient $b$ is $0 .$ These equations can be written in the form $a x^{2} + c = 0$ and have their own characteristics.

If the linear coefficient $b$ of a quadratic equation is $0,$ the equation is called a simple quadratic equation and can be written in the following form.

$a x^{2} + c = 0$

Rule

Number of Solutions of a Simple Quadratic Equation

This type of equation can be solved using inverse operations. Once $x^{2}$ is isolated, the equation can be written as $x^{2} = d,$ where $d = - \frac{c}{a} .$ The value of $d$ gives the number of solutions the equation has.

d > 0 : d = 0 : d < 0 : 2 real solutions 1 real solution 0 real solutions

Proof

The cases $d > 0,$ $d = 0,$ and $d < 0$ will be discussed one at a time.

$d > 0$

By taking square roots, the equation

x^{2} = d

can be rewritten.

x^{2} = d \Leftrightarrow x^{2} = d

In this case, because

d > 0,

the expression

d

is a real number. Therefore, the resulting equation can be solved.

x^{2} = d

$a^{2} = \pm a$

x = \pm d

StateSol

State solutions

x = d x = - d (I) (II)

It has been shown that if

d > 0,

the equation

x^{2} = d

has two real solutions which are

d

and

- d .

$d = 0$

d = 0,

then the equation

x^{2} = d

can be written as

x^{2} = 0 .

This equation can be solved for

x .

x^{2} = 0 \Leftrightarrow x \cdot x = 0

By using the Zero Product Property, it can be concluded that

x = 0 .

This is the only solution for the equation.

$d < 0$

Because the square of any real number is always greater than or equal to $0,$ if $d < 0$ the equation $x^{2} = d$ has no real solutions.

Example

Number of Solutions

Heichi is going on a trip with a friend. He wants to finish up his homework first, so he does not have to worry about it when he gets home.

He has been asked to determine the number of real solutions of three simple quadratic equations. Since Heichi only has a few minutes, he will determine the number of solutions without solving the equations. Help Heichi get ready for his trip!

Hint

Write the equations in the form $x^{2} = d .$ If $d > 0,$ the equation has two real solutions. If $d = 0,$ then the equation has one real solution. Finally, if $d < 0,$ the equation has no real solutions.

Solution

To start, the equation

- 4 x^{2} + 5 = 5

will be written in the form

x^{2} = d .

- 4 x^{2} + 5 = 5

▼

Solve for

x^{2}

SubEqn

$LHS - 5 = RHS - 5$

- 4 x^{2} = 0

DivEqn

$LHS / (- 4) = RHS / (- 4)$

x^{2} = 0

In this case

d

is equal to

0 .

Therefore, the equation

- 4 x^{2} + 5 = 5

has one real solution. By following a similar procedure, the other equations can be rewritten in the form

x^{2} = d .

Equation	Rewrite as $x^{2} = d$	Value of $d$	Number of Real Solutions
$- 4 x^{2} + 5 = 5$	$x^{2} = 0$	$d = 0$	One
$5 x^{2} - 125 = 0$	$x^{2} = 25$	$d = 25 \Rightarrow d > 0$	Two
$- 3 x^{2} - 27 = 0$	$x^{2} = - 9$	$d = - 9 \Rightarrow d < 0$	Zero

Pop Quiz

How Many Solutions?

Without solving the simple quadratic equations, determine the number of real solutions.

Discussion

Solving Quadratic Equations with Square Roots

Apart from determining the number of real solutions of a simple quadratic equation, most of the times it is important to calculate those solutions.

Simple quadratic equations are quadratic equations whose linear coefficient

b

is equal to

0 .

a x^{2} + c = 0

This type of equation can be solved using inverse operations, and two steps must be followed.

Isolate $x^{2} .$
Take square roots on both sides of the equation.

As an example, consider the equation

5 x^{2} - 500 = 0 .

Isolate

x^{2}

Inverse operations will be used to isolate

x^{2} .

Here,

500

will be added to both sides of the equation. Then, the left- and right-hand sides will be divided by

5 .

5 x^{2} - 500 = 0

AddEqn

$LHS + 500 = RHS + 500$

5 x^{2} = 500

DivEqn

$LHS / 5 = RHS / 5$

x^{2} = 100

Take Square Roots

Now that

x^{2}

has been isolated, it is necessary to undo the exponent. Exponents and radicals of the same index undo each other. Therefore, square roots undo powers of

2 .

x^{2} = 100

SqrtEqn

$LHS = RHS$

x^{2} = 100

$a^{2} = \pm a$

x = \pm 100

CalcRoot

Calculate root

x = \pm 10

StateSol

State solutions

x = 10 x = - 10

Note that the negative solution is also considered along with the principal root when solving the equation.

Example

Rational Solutions

Ali and Heichi are enjoying a ski vacation.

Heichi told Ali that he would pay for an extra hotel night if Ali could solve the following quadratic equation.

16 x^{2} + 15 = 40

Solve the equation and help Ali get an extra hotel night for free! Write the smallest solution first.

Hint

Start by isolating $x^{2} .$

Solution

In the quadratic equation, the linear coefficient is

0 .

Therefore, the given is a simple quadratic equation that can be solved by taking square roots. First,

x^{2}

needs to be isolated.

16 x^{2} + 15 = 40

▼

Solve for

x^{2}

SubEqn

$LHS - 15 = RHS - 15$

16 x^{2} = 25

DivEqn

$LHS / 16 = RHS / 16$

x^{2} = \frac{2 5}{1 6}

Now that

x^{2}

has been isolated, square roots can be taken on the left- and the right-hand sides. Both the principal root and the negative solution will be considered.

x^{2} = \frac{2 5}{1 6}

SqrtEqn

$LHS = RHS$

x^{2} = \frac{2 5}{1 6}

$a^{2} = \pm a$

x = \pm \frac{2 5}{1 6}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

x = \pm \frac{2 5}{1 6}

CalcRoot

Calculate root

x = \pm \frac{5}{4}

StateSol

State solutions

x = \frac{5}{4} x = - \frac{5}{4} (I) (II)

The equation has two real solutions and both of them are rational.

Example

Irrational Solutions

Dominika and Magdalena are enjoying a vacation at a beach resort.

Dominika told Magdalena that she would pay for an extra hotel night if Magdalena could solve the following quadratic equation.

- 2 x^{2} + 7 = 4

Solve the equation and help Magdalena get an extra hotel night for free! Round the solutions to three significant figures and write the smallest solution first.

Hint

Start by isolating $x^{2} .$

Solution

In the quadratic equation, the linear coefficient is

0 .

Therefore, the given is a simple quadratic equation and therefore can be solved by taking square roots. First,

x^{2}

needs to be isolated.

- 2 x^{2} + 7 = 4

▼

Solve for

x^{2}

SubEqn

$LHS - 7 = RHS - 7$

- 2 x^{2} = - 3

DivEqn

$LHS / (- 2) = RHS / (- 2)$

x^{2} = \frac{- 3}{- 2}

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

x^{2} = \frac{3}{2}

Now that

x^{2}

has been isolated, square roots can be taken on the left- and the right-hand sides. Both the principal root and the negative solution will be considered.

x^{2} = \frac{3}{2}

SqrtEqn

$LHS = RHS$

x^{2} = \frac{3}{2}

$a^{2} = \pm a$

x = \pm \frac{3}{2}

StateSol

State solutions

x = \frac{3}{2} x = - \frac{3}{2} (I) (II)

$(I), (II):$ Use a calculator

x = 1.224744 \dots x = - 1.224744 \dots

$(I), (II):$ Round to $3$ significant digit(s)

x \approx 1.22 x \approx - 1.22

The equation has two real solutions,both of which are irrational.

Pop Quiz

Solving Simple Quadratic Equations

Solve the following simple quadratic equations by taking square roots. If necessary, round the solutions to two decimal places.

Example

Solving Other Type of Quadratic Equation

Jordan is representing North High School in an algebra competition.

She has been challenged with a quadratic equation that is a bit more complicated than a simple quadratic equation.

- 2 (x - 5)^{2} + 2 = 0

Jordan realizes that the equation can be solved by taking square roots. Help her solve the equation! Write the smallest solution first.

Hint

Start by isolating $(x - 5)^{2} .$

Solution

To solve the given equation, the expression

(x - 5)^{2}

will be isolated first. Then square roots will be taken on both sides.

- 2 (x - 5)^{2} + 2 = 0

▼

Solve for

(x - 5)^{2}

SubEqn

$LHS - 2 = RHS - 2$

- 2 (x - 5)^{2} = - 2

DivEqn

$LHS / (- 2) = RHS / (- 2)$

(x - 5)^{2} = 1

Now that

(x - 5)^{2}

has been isolated, square roots can be taken on the left- and the right-hand sides. Both the principal root and the negative solution will be considered.

(x - 5)^{2} = 1

SqrtEqn

$LHS = RHS$

(x - 5)^{2} = 1

$a^{2} = \pm a$

x - 5 = \pm 1

CalcRoot

Calculate root

x - 5 = \pm 1

StateSol

State solutions

x - 5 = 1 x - 5 = - 1 (I) (II)

$(I), (II):$ $LHS + 5 = RHS + 5$

x = 6 x = 4

Closure

A Process Called Completing the Square

The quadratic equation given in the last example had a specific format.

Equation : Format : - 2 (x - 5)^{2} + 2 = 0 a (x - h)^{2} + k = 0

It is worth noting that all quadratic equations can be written in this format by a process called completing the square.

Equation	Rewrite
$- x^{2} + 4 = 0$	$- 1 (x - 0)^{2} + 4 = 0$
$3 x^{2} - 6 x + 5 = 0$	$3 (x - 1)^{2} + 2 = 0$
$5 x^{2} = 3 x + 1$	$5 (x - \frac{3}{1 0})^{2} + (- \frac{2 9}{2 0}) = 0$
$4 x^{2} + 4 x = - 2$	$4 (x - (- \frac{1}{2}))^{2} + 1 = 0$
$4 x^{2} + 4 x = - 2$	$4 (x - (- \frac{1}{2}))^{2} + 1 = 0$
$10 x^{2} + 160 = 80 x$	$10 (x - 4)^{2} + 0 = 0$

For more information, read about the method of completing the square.

Solving Quadratic Equations Using Square Roots

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Return to lesson.

Solving Quadratic Equations Using Square Roots

Catch-Up and Review

Proof

d>0

d=0

d<0

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution