Solve the quadratic equation. $2 ( x + 8 ) ^ { 2 } - 5 = 3$ Use the following property: For any real number $D$ , $x ^ { 2 } = a$ if and only if $x = \sqrt { a }$ or $x = - \sqrt { a }$ .
A)x = -10 or x = -9
B)x = 0
C)x = -6 or x = -10
D)x = 1 or x = -3
E)x = -1 or x = -13

Question

Quizplus · Accepted Answer

First, simplify the left side of the equation:

$2(x+8)^2 - 5 = 3$

$2(x^2 + 16x + 64) - 5 = 3$

$2x^2 + 32x + 123 = 0$

Next, we can use the quadratic formula to solve for $x$:

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

In this case, $a=2$, $b=32$, and $c=123$. Substituting into the formula gives:

$x = \frac{-32 \pm \sqrt{32^2 - 4(2)(123)}}{2(2)}$

$x = \frac{-32 \pm \sqrt{256}}{4}$

$x = \frac{-32 \pm 16}{4}$

$x = -6$ or $x=-10$

Therefore, the answer is C.

Solve the Quadratic Equation 2(x+8)2−5=32 ( x + 8 ) ^ { 2 } - 5 = 32(x+8)2−5=3

Solve the Quadratic Equation $2 ( x + 8 ) ^ { 2 } - 5 = 3$