Question

question 9 of 10
complete the square to solve the equation below.
$x^2 + 2x - 7 = 17$

a. $x = -1, x = 4$

b. $x = 5, x = -4$

c. $x = -6, x = 4$

d. $x = -1, x = 3$

Explanation:

Step1: Move constant to right

First, we move the constant term \(-7\) to the right side of the equation by adding \(7\) to both sides. So the equation \(x^{2}+2x - 7=17\) becomes \(x^{2}+2x=17 + 7\), which simplifies to \(x^{2}+2x=24\).

Step2: Complete the square

To complete the square for the quadratic expression \(x^{2}+2x\), we take half of the coefficient of \(x\) (which is \(2\)), square it. Half of \(2\) is \(1\), and squaring \(1\) gives \(1\). We add this to both sides of the equation: \(x^{2}+2x + 1=24 + 1\). The left side is now a perfect square trinomial, which factors as \((x + 1)^{2}\), so we have \((x + 1)^{2}=25\).

Step3: Solve for x

Take the square root of both sides: \(x + 1=\pm\sqrt{25}=\pm5\). Now we solve for \(x\) in two cases.

• Case 1: \(x + 1 = 5\), then \(x=5 - 1=4\).
• Case 2: \(x + 1=-5\), then \(x=-5 - 1=-6\).

So the solutions are \(x = 4\) and \(x=-6\).

Answer:

C. \(x = - 6,x = 4\)