body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Glencoe Algebra 2, 2012

MH

McGraw Hill Glencoe Algebra 2, 2012 View details

Study Guide and Review

Continue to next subchapter

Exercise 63 Page 650

To solve the equation $a x^{2} + b x + c = 0,$ use the Quadratic Formula.

$(4, 0)$

Practice makes perfect

We will solve the given system of equations using the Substitution Method.

{y + x^{2} = 4 x y + 4 x = 16 (I) (II)

Note that neither of the variables is isolated in either equation, so we need to start by isolating

y

in Equation (I).

{y + x^{2} = 4 x y + 4 x = 16 \Leftrightarrow {y = - x^{2} + 4 x y + 4 x = 16

The

y -

variable is isolated in Equation (I). This allows us to substitute its value

- x^{2} + 4 x

for

y

in Equation (II).

{y = - x^{2} + 4 x y + 4 x = 16

$(II):$ $y = - x^{2} + 4 x$

{y = - x^{2} + 4 x - x^{2} + 4 x + 4 x = 16

▼

(II):

Simplify

$(II):$ Add terms

{y = - x^{2} + 4 x - x^{2} + 8 x = 16

$(II):$ $LHS - 16 = RHS - 16$

{y = - x^{2} + 4 x - x^{2} + 8 x - 16 = 0

Notice that in Equation (II), we have a quadratic equation in terms of only the

x -

variable.

- x^{2} + 8 x - 16 = 0 \Leftrightarrow - 1 x^{2} + 8 x + (- 16) = 0

We can substitute

a = - 1,

b = 8,

and

c = - 16

into the Quadratic Formula.

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

SubstituteValues

Substitute values

x = \frac{- 8 \pm 8 ^{2} - 4 ( - 1 ) ( - 1 6 )}{2 ( - 1 )}

▼

Simplify right-hand side

Calculate power

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

MultNegNegOnePar

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

x = \frac{- 8 \pm 6 4 + 4 ( - 1 6 )}{2 ( - 1 )}

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

x = \frac{- 8 \pm 6 4 - 6 4}{- 2}

Subtract term

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

Calculate root

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

Add and subtract terms

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

Calculate quotient

x = 4

Now, consider Equation (I).

y = - x^{2} + 4 x

We can substitute

4

for

x

in the above equation to find the value for

y .

y = - x^{2} + 4 x

$x = 4$

y = - 4^{2} + 4 (4)

▼

Simplify right-hand side

Calculate power and product

y = - 16 + 16

Add terms

y = 0

We found that

y = 0

when

x = 4 .

Thus, the solution of the system is

(4, 0) .

Checking Our Answer

Checking the answer

We can check our answers by substituting the solution into both equations. If they produce true statements, our solution is correct. We will substitute

4

and

0

for

x

and

y,

respectively, in Equation (I) and Equation (II).

{y + x^{2} = 4 x y + 4 x = 16

$(I), (II):$ $x = 4$ , $y = 0$

{0 + 4^{2} = ? 4 (4) 0 + 4 (4) = ? 16

▼

Simplify

$(I):$ Calculate power

{0 + 16 = ? 4 (4) 0 + 4 (4) = ? 16

$(I), (II):$ Multiply

{0 + 16 = ? 16 0 + 16 = ? 16

$(I), (II):$ Add terms

{16 = 16 ✓ 16 = 16 ✓

Since both equations produced true statements, the solution

(4, 0)

is correct.

Subchapter links

Exercises p.646-650