Unlocking Solving Quadratic Equations Using the Quadratic Formula

There are several methods for solving quadratic equations. In this lesson, the Quadratic Formula will be presented, proven, and used. Additionally, a method for determining the number of solutions without actually solving the equation will be presented.

Catch-Up and Review

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

Square Root
Quadratic Equation
Standard Form of a Quadratic Function
Completing the Square

Try a few practice exercises as a warm-up!

a Simplify the numeric expression

4^{2} - 4 (1) (2)

by using the properties of square roots to remove the perfect-square factor.

b Consider the quadratic function

y = (x + 2)^{2} - 1 .

Of the following expressions, which represents the same function written in standard form?

c Identify the coefficient

b

of the quadratic equation

x - 5 = 2 (x - 3)^{2}

when written in standard form

a x^{2} + b x + c = 0 .

Challenge

Is There a Solution?

Magdalena and Diego, both huge fans of statistics, went camping to bond under the stars and talk stats. However, they realize that bears are in the area. They need to hang their food basket from a branch $15$ feet above the ground. Diego figures he can throw a stone with a rope attached to it over the branch. As Diego winds up, Magdalena sheepishly snickers, "No way that works."

Throwing a stone on a branch during camping

In wondering if Diego's throw will be a success, consider the following quadratic function that models the height, in air, of the stone's location after

t

seconds of being thrown.

h (t) = 5 (- 3 t^{2} + 5 t + 1)

Magdalena also wonders what quadratic equation represents this scenario. Help her find it. Then, without solving the equation, determine whether it is even possible to know if the stone will reach the branch.

Discussion

The Quadratic Formula

Besides graphing, using square roots, factoring, and completing the square, there is another method for solving a quadratic equation. This method consists of using the Quadratic Formula. Check out how to derive the formula by completing the square!

Rule

Quadratic Formula

The Quadratic Formula can be used to solve a quadratic equation written in standard form $a x^{2} + b x + c = 0 .$

$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Proof

The Quadratic Formula can be derived by completing the square given the standard form of the quadratic equation

a x^{2} + b x + c = 0 .

This method will be used to isolate the

x -

variable. To complete the square, there are five steps to follow.

Factor Out the Coefficient of

x^{2}

It is easier to complete the square when the quadratic expression is written in the form

x^{2} + b x + c .

Therefore, the coefficient

a

should be factored out.

a x^{2} + b x + c = 0 ⇕ a (x^{2} + \frac{b}{a} x + \frac{c}{a}) = 0

Since the equation is quadratic, the coefficient

a

is not equal to

0 .

Therefore, both sides of the equation can be divided by

a .

a (x^{2} + \frac{b}{a} x + \frac{c}{a}) = 0 ⇕ x^{2} + \frac{b}{a} x + \frac{c}{a} = 0

Identify the Constant Needed to Complete the Square

The next step is to rewrite the equation by moving the existing constant to the right-hand side. To do so,

\frac{c}{a}

will be subtracted from both sides of the equation.

x^{2} + \frac{b}{a} x + \frac{c}{a} = 0 ⇕ x^{2} + \frac{b}{a} x = - \frac{c}{a}

The constant needed to complete the square can now be identified by focusing on the

x -

term, while ignoring the rest. One way to find this constant is by squaring half the coefficient of the

x -

term, which in this case is

\frac{b}{a} .

(\frac{\frac{b}{a}}{2})^{2}

DivFracD

$\frac{a / c}{b} = \frac{a}{b \cdot c}$

(\frac{b}{2 a})^{2}

Note that leaving the constant as a power makes the next steps easier to perform.

Complete the Square

The square can now be completed by adding the

constant

found in Step

2

to both sides of the equation.

x^{2} + \frac{b}{a} x = - \frac{c}{a} ⇕ Perfect Square Trinomial x^{2} + \frac{b}{a} x + (\frac{b}{2 a})^{2} = Constant - \frac{c}{a} + (\frac{b}{2 a})^{2}

The first three terms form a perfect square trinomial, which can be factored as the square of a binomial. The other two terms do not contain the variable

x .

Therefore, their value is constant.

Factor the Perfect Square Trinomial

The perfect square trinomial can now be factored and rewritten as the square of a binomial.

x^{2} + \frac{b}{a} x + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

x^{2} + 2 (\frac{b}{2 a}) x + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

CommutativePropMult

Commutative Property of Multiplication

x^{2} + 2 x (\frac{b}{2 a}) + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

FacPosPerfectSquare

$a^{2} + 2 a b + b^{2} = (a + b)^{2}$

(x + \frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

The process of completing the square is now finished.

Simplify the Equation

Finally, the right-hand side of the equation can be simplified.

(x + \frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

▼

Simplify right-hand side

CommutativePropAdd

Commutative Property of Addition

(x + \frac{b}{2 a})^{2} = (\frac{b}{2 a})^{2} - \frac{c}{a}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{( 2 a ) ^{2}} - \frac{c}{a}

PowProdII

$(a b)^{m} = a^{m} b^{m}$

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{c}{a}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 4 a}{b \cdot 4 a}$

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{c \cdot 4 a}{a \cdot 4 a}

CommutativePropMult

Commutative Property of Multiplication

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{4 a c}{4 a \cdot a}

ProdToPowTwoFac

$a \cdot a = a^{2}$

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{4 a c}{4 a ^{2}}

SubFrac

Subtract fractions

(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}

Now, there is only one

x -

term. To isolate

x,

it is necessary to take square roots on both sides of the equation. This results in both a positive and a negative term on the right-hand side.

(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}} ⇕ x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{4 a ^{2}}

Now, the equation can be further simplified to isolate

x .

x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{4 a ^{2}}

▼

Solve for

x

SqrtQuot

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

x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{4 a ^{2}}

SqrtProd

$a \cdot b = a \cdot b$

x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{2 a ^{2}}

SqrtPowToNumber

$a^{2} = a$

x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{2 a}

SubEqn

$LHS - \frac{b}{2 a} = RHS - \frac{b}{2 a}$

x = - \frac{b}{2 a} \pm \frac{b ^{2} - 4 a c}{2 a}

MoveNegFracToNum

Put minus sign in numerator

x = \frac{- b}{2 a} \pm \frac{b ^{2} - 4 a c}{2 a}

AddSubFrac

Add and subtract fractions

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Finally, the Quadratic Formula has been obtained.

$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Example

Solving a Quadratic Equation Using the Quadratic Formula

Magdalena will sell lottery tickets as a fundraiser to support paralympic athletes. The total profit

p (x)

depends on the price

x

of a ticket and can be modeled by using the following quadratic equation.

p (x) = - 2 x^{2} + 32 x + 104

Magdalena wants to raise at least

$ 200 .

However, she has not yet set the price of each lottery ticket. Help Magdalena find the smallest amount that can be charged per ticket and still make a profit of at least

$ 200 .

Round the price to the nearest whole dollar (the dollar sign is not necessary).

Hint

Since the profit should be at least $$ 200,$ let $p (x)$ be equal to $200 .$ Then, rewrite the quadratic equation in standard form. The equation can be solved using the Quadratic Formula.

Solution

It is given that the profit for the fundraiser should be at least

$ 200 .

Consider the given quadratic equation, which models the profit, and substitute

200

for

p (x) .

Then, rewrite the obtained equation in standard form.

p (x) = - 2 x^{2} + 32 x + 104

Substitute

$p (x) = 200$

200 = - 2 x^{2} + 32 x + 104

SubEqn

$LHS - 200 = RHS - 200$

0 = - 2 x^{2} + 32 x - 96

RearrangeEqn

Rearrange equation

- 2 x^{2} + 32 x - 96 = 0

Now, all of the coefficients in the standard form

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

can be determined.

- 2 x^{2} + 32 x - 96 = 0 ⇕ - 2 x^{2} + 32 x + (- 96) = 0

Therefore,

a = - 2,

b = 32,

and

c = - 96 .

The obtained equation will be solved using the Quadratic Formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The values of

a,

b,

and

c

will now be substituted into the formula. Find

x

by evaluating the right-hand side of the formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- 3 2 \pm 3 2 ^{2} - 4 ( - 2 ) ( - 9 6 )}{2 ( - 2 )}

▼

Evaluate right-hand side

CalcPow

Calculate power

x = \frac{- 3 2 \pm 1 0 2 4 - 4 ( - 2 ) ( - 9 6 )}{2 ( - 2 )}

MultPosNeg

$a (- b) = - a \cdot b$

x = \frac{- 3 2 \pm 1 0 2 4 - ( - 8 ) ( - 9 6 )}{- 4}

MultNegNegTwoPar

$(- a) (- b) = a \cdot b$

x = \frac{- 3 2 \pm 1 0 2 4 - 7 6 8}{- 4}

SubTerm

Subtract term

x = \frac{- 3 2 \pm 2 5 6}{- 4}

CalcRoot

Calculate root

x = \frac{- 3 2 \pm 1 6}{- 4}

Using the Quadratic Formula, it was obtained that the solutions for the equation are

x = \frac{- 3 2 \pm 1 6}{- 4} .

Finally, both solutions can be evaluated using a table.

$x = \frac{- 3 2 \pm 1 6}{- 4}$
$x = \frac{- 3 2 + 1 6}{- 4}$	$x = \frac{- 3 2 - 1 6}{- 4}$
$x = \frac{- 1 6}{- 4}$	$x = \frac{- 4 8}{- 4}$
$x = 4$	$x = 12$

Since Magdalena wants the tickets to be as cheap as possible while making a profit of at least $$ 200,$ the price each ticket should be $$ 4 .$

Example

Solving a Quadratic Equation Not in Standard Form

A fire nozzle attached to a hose is a device used by firefighters to extinguish fires. Consider a firefighter who is aiming water to extinguish a fire on the third floor of a building. The base of the fire is situated $22$ feet above the ground.

Firefighter delivering water into the window of a building

The stream of water delivered from the fire nozzle can be modeled by the following quadratic function.

h (x) = - 0.008 x (x - 100) + 4

In this equation,

x

is the horizontal distance from the firefighter and

h (x)

is the height of the water stream. Both

x

and

h (x)

are measured in feet. Knowing that the water stream's peak is

2

feet above the base of the fire, what is the horizontal distance from the firefighter to the peak of the water stream?

Hint

What is the height of the water stream's peak? Write a quadratic equation and solve it using the Quadratic Formula.

Solution

Consider the given situation on a coordinate plane and assume that the firefighter is standing on the

y -

axis.

Since the water stream's peak is

2

feet above the fire's base, whose height is

22

feet, its height

h (x)

2 + 22 = 24

feet. This height will now be substituted into the equation of the given quadratic function to calculate the desired distance.

h (x) = - 0.008 x (x - 100) + 4 ↓ 24 = - 0.008 x (x - 100) + 4

The obtained quadratic equation can be solved using the Quadratic Formula. To do so, the equation must first be rewritten in standard form.

24 = - 0.008 x (x - 100) + 4

▼

Rewrite

Distr

Distribute $- 0.008 x$

24 = - 0.008 x^{2} + 0.8 x + 4

SubEqn

$LHS - 24 = RHS - 24$

0 = - 0.008 x^{2} + 0.8 x - 20

RearrangeEqn

Rearrange equation

- 0.008 x^{2} + 0.8 x - 20 = 0

Next, the coefficients

a,

b,

and

c

can be identified.

- 0.008 x^{2} + 0.8 x - 20 = 0 ⇕ - 0.008 x^{2} + 0.8 x + (- 20) = 0

Finally, these values will be substituted into the Quadratic Formula to solve the equation for

x .

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- 0 . 8 \pm 0 . 8 ^{2} - 4 ( - 0 . 0 0 8 ) ( - 2 0 )}{2 ( - 0 . 0 0 8 )}

▼

Evaluate right-hand side

CalcPow

Calculate power

x = \frac{- 0 . 8 \pm 0 . 6 4 - 4 ( - 0 . 0 0 8 ) ( - 2 0 )}{2 ( - 0 . 0 0 8 )}

MultPosNeg

$a (- b) = - a \cdot b$

x = \frac{- 0 . 8 \pm 0 . 6 4 - ( - 0 . 0 3 2 ) ( - 2 0 )}{- 0 . 0 1 6}

MultNegNegTwoPar

$(- a) (- b) = a \cdot b$

x = \frac{- 0 . 8 \pm 0 . 6 4 - 0 . 6 4}{- 0 . 0 1 6}

SubTerm

Subtract term

x = \frac{- 0 . 8 \pm 0}{- 0 . 0 1 6}

CalcRoot

Calculate root

x = \frac{- 0 . 8 \pm 0}{- 0 . 0 1 6}

AddSubTerms

Add and subtract terms

x = \frac{- 0 . 8}{- 0 . 0 1 6}

DivNegNeg

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

x = \frac{0 . 8}{0 . 0 1 6}

CalcQuot

Calculate quotient

x = 50

It has been found that this equation has exactly one solution,

x = 50 .

Therefore, it can be said that the firefighter is standing at a horizontal distance of

50

feet from the water stream's peak.

Pop Quiz

Solving a Quadratic Equation Using the Quadratic Formula

Solve the quadratic equations by using the Quadratic Formula. If necessary, round the answer to $2$ decimal places.

Discussion

Discriminant of a Quadratic Equation

In general, quadratic equations have two, one or no real solutions. Before solving a quadratic equation, the number of real solutions can be determined by using the discriminant.

Concept

Discriminant

In the Quadratic Formula, the expression $b^{2} - 4 a c,$ which is under the radical symbol, is called the discriminant.

$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

A quadratic equation can have two, one, or no real solutions. Since the discriminant is under the radical symbol, its value determines the number of real solutions of a quadratic equation.

Value of the Discriminant	Number of Real Solutions
$b^{2} - 4 a c > 0$	$2$
$b^{2} - 4 a c = 0$	$1$
$b^{2} - 4 a c < 0$	$0$

Moreover, the discriminant determines the number of $x -$ intercepts of the graph of the related quadratic function.

The number of $x$-intercepts of the graph of a quadratic function

Example

Determining Whether There is a Solution Without Solving

A farmer wants to build a fence around a vegetable garden. To make it simple, the farmer will build it in the shape of a rectangle. The farmer has enough wood to build a fence the length of

800

feet, including the gate.

A part of the land for a vegetable garden

The area of the vegetable garden changes as the side lengths of the rectangle change. The farmer wants the garden's area to be at least

50000

square feet. Will he need to buy more wood to achieve this goal?

Hint

Let $x$ denote the length of one side of the rectangle. Then, use the fact that the length of the fence represents the perimeter of the rectangle. All things considered, how can the area of the rectangle be calculated?

Solution

Let

x

and

y

be the side lengths of the rectangle that represents the vegetable garden. The length of the fence is the perimeter

P

of the rectangle, which is the sum of all four side lengths.

P = x + x + y + y ⇕ P = 2 x + 2 y

It is given that the perimeter of the rectangle — how much wood the farmer has — is

800

feet. By substituting

800

for

P,

the equation can be solved for one of the side lengths, such as

y .

P = 2 x + 2 y

Substitute

$P = 800$

800 = 2 x + 2 y

▼

Solve for

y

SubEqn

$LHS - 2 x = RHS - 2 x$

800 - 2 x = 2 y

DivEqn

$LHS / 2 = RHS / 2$

\frac{8 0 0 - 2 x}{2} = y

WriteDiffFrac

Write as a difference of fractions

\frac{8 0 0}{2} - \frac{2 x}{2} = y

MovePartNumRight

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

\frac{8 0 0}{2} - \frac{2}{2} x = y

CalcQuot

Calculate quotient

400 - 1 x = y

IdPropMult

Identity Property of Multiplication

400 - x = y

RearrangeEqn

Rearrange equation

y = 400 - x

The side lengths can now be placed in the diagram. It can be arbitrarily assumed that the length of the horizontal side is

x .

Keep in mind that both

x

and

400 - x

are measured in feet.

Next, the area of the rectangle will be calculated in terms of

x .

The area

A

of a rectangle is the product of the rectangle's length and width.

A = x (400 - x)

The obtained formula for

A

is represented by a quadratic function. It is given that the farmer's desired area should be at least

50000

square feet. Therefore, this number will be substituted for

A

in the formula.

A = x (400 - x) ↓ 50000 = x (400 - x)

The above is a quadratic equation that is not written in standard form. Hence, the equation will be rewritten to determine the number of solutions. Determining the number of solutions will help find if a value for

x

exists so that the area of the rectangle is

50000

square feet.

50000 = x (400 - x)

▼

Rewrite

Distr

Distribute $x$

50000 = 400 x - x^{2}

SubEqn

$LHS - 50000 = RHS - 50000$

0 = 400 x - x^{2} - 50000

CommutativePropAdd

Commutative Property of Addition

0 = - x^{2} + 400 x - 50000

RearrangeEqn

Rearrange equation

- x^{2} + 400 x - 50000 = 0

The equation is now in standard form. This means that the number of solutions can be determined using the discriminant. Next, the coefficients

a,

b,

and

c

need to be identified.

- x^{2} + 400 x - 50000 = 0 ⇕ - 1 x^{2} + 400 x + (- 50000) = 0

The variables

a,

b,

and

c

can then be substituted into the discriminant

b^{2} - 4 a c .

b^{2} - 4 a c

SubstituteValues

Substitute values

400^{2} - 4 (- 1) (- 50000)

▼

Evaluate

CalcPow

Calculate power

160000 - 4 (- 1) (- 50000)

MultPosNeg

$a (- b) = - a \cdot b$

160000 - (- 4) (- 50000)

MultNegNegOnePar

$- a (- b) = a \cdot b$

160000 - 200000

SubTerm

Subtract term

- 40000

Since the discriminant is less than

0,

there are no solutions to the equation. Therefore, the farmer will not be able to build a fence so that the area of the vegetable garden is

50000

square feet. This means that he will need to buy more wood. Good thing he did the math before starting to construct the fence.

Pop Quiz

Determining the Number of Real Solutions of a Quadratic Equation

Without solving the quadratic equations, use the discriminant to determine the number of real solutions.

Determine the number of real-number solutions of a quadratic equation

Closure

Determining the Number of Solutions of a Quadratic Equation Without Solving

The challenge presented at the beginning of this lesson asked if the stone thrown by Diego will reach, over some point in time, a branch located $15$ feet above the ground.

The height, in feet, of the stone thrown by Diego is modeled by the following quadratic function.

h (t) = 5 (- 3 t^{2} + 5 t + 1)

Will the stone reach the branch? There is no need to solve any equation to answer the question.

Hint

Substitute $15$ for $h (t)$ and identify the discriminant of the resulting quadratic equation.

Solution

Here,

t

represents the number of seconds that passed since Diego threw the stone. Since the branch is situated

15

feet above the ground, this number can be substituted for

h (t)

into the formula.

h (t) = 5 (- 3 t^{2} + 5 t + 1) ↓ 15 = 5 (- 3 t^{2} + 5 t + 1)

Although this quadratic equation can be solved using the Quadratic Formula, it is sufficient to find whether the equation has any solutions. To do so, the discriminant can be calculated. To calculate the discriminant, the equation needs to be rewritten in standard form.

15 = 5 (- 3 t^{2} + 5 t + 1)

▼

Rewrite

Distr

Distribute $5$

15 = - 15 t^{2} + 25 t + 5

SubEqn

$LHS - 15 = RHS - 15$

0 = - 15 t^{2} + 25 t - 10

RearrangeEqn

Rearrange equation

- 15 t^{2} + 25 t - 10 = 0

Now, identify the coefficients

a,

b,

and

c

in the obtained equation.

- 15 t^{2} + 25 t - 10 = 0 ⇕ - 15 t^{2} + 25 t + (- 10) = 0

Finally, the values of the coefficients will be substituted into the discriminant.

b^{2} - 4 a c

SubstituteValues

Substitute values

25^{2} - 4 (- 15) (- 10)

▼

Evaluate

CalcPow

Calculate power

625 - 4 (- 15) (- 10)

MultPosNeg

$a (- b) = - a \cdot b$

625 - (- 60) (- 10)

MultNegNegOnePar

$- a (- b) = a \cdot b$

625 - 600

SubTerm

Subtract term

25

Because the value of the discriminant is greater than

0,

there are two solutions of the equation. Therefore, Diego and Magdalena know that the stone will reach the desired branch at two points in time.

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Catch-Up and Review

Proof

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution