Question

question 10 of 10
complete the square to solve the equation below.
check all that apply.
$x^2 + 8x - 2 = 18$

a. $x = 2$
b. $x = -10$
c. $x = -3$
d. $x = -2$

Explanation:

Step1: Move constant to right

First, we move the constant term \(-2\) to the right side of the equation by adding \(2\) to both sides. The equation \(x^{2}+8x - 2=18\) becomes \(x^{2}+8x=18 + 2\), so \(x^{2}+8x=20\).

Step2: Complete the square

To complete the square for the quadratic expression \(x^{2}+8x\), we take half of the coefficient of \(x\) (which is \(8\)), square it, and add it to both sides. Half of \(8\) is \(4\), and squaring \(4\) gives \(16\). So we add \(16\) to both sides of the equation \(x^{2}+8x = 20\):
\(x^{2}+8x+16=20 + 16\)
The left side can be written as a perfect square: \((x + 4)^{2}=36\)

Step3: Solve for x

Take the square root of both sides: \(x+4=\pm\sqrt{36}=\pm6\)
Now we solve for \(x\) in two cases:

• Case 1: \(x + 4=6\), then \(x=6 - 4=2\)
• Case 2: \(x + 4=-6\), then \(x=-6 - 4=-10\)

Answer:

A. \(x = 2\), B. \(x=-10\)