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McGraw Hill Glencoe Algebra 1, 2012 View details

5. Solving Quadratic Equations by Using the Quadratic Formula

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Exercise 41 Page 587

a We are given a quadratic function that estimates the percent of U.S. households with high-speed Internet

n

years after

1990 .

h = - 0.2 n^{2} + 7.2 n + 1.5

To determine when

20 %

of the U.S. population will have high-speed Internet, we need to substitute

20

for

h

into the given quadratic function.

20 = - 0.2 n^{2} + 7.2 n + 1.5

Let's rewrite the equation in standard form.

20 = - 0.2 n^{2} + 7.2 n + 1.5

SubEqn

$LHS - 20 = RHS - 20$

0 = - 0.2 n^{2} + 7.2 n - 18.5

RearrangeEqn

Rearrange equation

- 0.2 n^{2} + 7.2 n - 18.5 = 0

Now, we will solve it by using the Quadratic Formula. We first need to identify the values of

a,

b,

and

c .

- 0.2 n^{2} + 7.2 n - 18.5 = 0 ⇕ - 0.2 n^{2} + 7.2 n + (- 18.5) = 0

We see that

a = - 0.2,

b = 7.2,

and

c = - 18.5 .

Let's substitute these values into the Quadratic Formula.

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

SubstituteValues

Substitute values

n = \frac{- 7 . 2 \pm ( 7 . 2 ) ^{2} - 4 ( - 0 . 2 ) ( - 1 8 . 5 )}{2 ( - 0 . 2 )}

▼

Simplify right-hand side

CalcPow

Calculate power

n = \frac{- 7 . 2 \pm 5 1 . 8 4 - 4 ( - 0 . 2 ) ( - 1 8 . 5 )}{2 ( - 0 . 2 )}

Multiply

n = \frac{- 7 . 2 \pm 5 1 . 8 4 - 4 ( - 0 . 2 ) ( - 1 8 . 5 )}{- 0 . 4}

MultNegNegTwoPar

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

n = \frac{- 7 . 2 \pm 5 1 . 8 4 - 4 ( 3 . 7 )}{- 0 . 4}

MultNegPos

$(- a) b = - a b$

n = \frac{- 7 . 2 \pm 5 1 . 8 4 - 1 4 . 8}{- 0 . 4}

SubTerm

Subtract term

n = \frac{- 7 . 2 \pm 3 7 . 0 4}{- 0 . 4}

The solutions for this equation are

n = \frac{- 7 . 2 \pm 3 7 . 0 4}{- 0 . 4} .

Let's separate them into the positive and negative cases.

$n = \frac{- 7 . 2 \pm 3 7 . 0 4}{- 0 . 4}$
$n_{1} = \frac{- 7 . 2 + 3 7 . 0 4}{- 0 . 4}$	$n_{2} = \frac{- 7 . 2 - 3 7 . 0 4}{- 0 . 4}$
$n_{1} = \frac{7 . 2}{0 . 4} - \frac{3 7 . 0 4}{0 . 4}$	$n_{2} = \frac{7 . 2}{0 . 4} + \frac{3 7 . 0 4}{0 . 4}$
$n_{1} \approx 3$	$n_{2} \approx 33$

Using the Quadratic Formula, we found that the solutions of the given equation are

n_{1} \approx 3

and

x_{2} \approx 33 .

Since

n

is the number of years since

1990,

1993

and

2023,

20 %

of the U.S. population will have high-speed Internet.

1990 + 3 = 1993 1990 + 33 = 2023

b Since we are given a quadratic equation with a negative leading coefficient, the function has an maximum point.

h = - 0.2 n^{2} + 7.2 n + 1.5

Let's find this point to check if this quadratic equation is a good model. To do so, we need to identify

a -,

b -,

and

c -

values of the related quadratic function.

f (x) = - 0.2 x^{2} + 7.2 x + 1.5

We see that

a = - 0.2,

b = 7.2,

and

c = 1.5 .

Recall that the maximum point is the vertex. We can write the expression for the vertex by stating the

x -

and

f -

coordinates in terms of

a

and

b .

Vertex : (- \frac{b}{2 a}, f (- \frac{b}{2 a}))

We can now find the

x -

coordinate of the vertex by substituting the values.

x = - \frac{b}{2 a}

SubstituteII

$a = - 0.2$ , $b = 7.2$

x = - \frac{7 . 2}{2 ( - 0 . 2 )}

▼

Simplify right-hand side

MultPosNeg

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

x = - \frac{7 . 2}{- 0 . 4}

RemoveNegFracAndDenom

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

x = \frac{7 . 2}{0 . 4}

CalcQuot

Calculate quotient

x = 18

To find the second coordinate, we need to substitute

18

for

x

in the related function.

f = - 0.2 n^{2} + 7.2 n + 1.5

Substitute

$x = 18$

f = - 0.2 (18)^{2} + 7.2 (18) + 1.5

▼

Simplify right-hand side

CalcPow

Calculate power

f = - 0.2 (324) + 7.2 (18) + 1.5

Multiply

f = - 64.8 + 129.6 + 1.5

AddTerms

Add terms

f = 66.3

RoundInt

Round to nearest integer

f \approx 66

The vertex is

(18, 66) .

Hence, the maximum

h -

value is about

66,

meaning that only

66 %

of the population will have high-speed Internet. As a result, this quadratic equation is not a good model.

Subchapter links

Check Your Understanding p.587

Practice and Problem Solving p.587-588

H.O.T. Problems p.588

Standardized Test Practice p.589

Spiral Review p.589

Skills Review p.589