Question

solve by completing the square.
$x^2 - 4x - 31 = -10$
select the correct response:
$4, -10$
$-3 + \sqrt{59}, -3 - \sqrt{59}$
$7, -3$
$-10 + \sqrt{27}, -10 - \sqrt{27}$

Explanation:

Step1: Isolate the quadratic and linear terms

First, we move the constant term to the right side of the equation. The original equation is \(x^{2}-4x - 31=- 10\). Add 31 to both sides:
\(x^{2}-4x=-10 + 31\)
\(x^{2}-4x=21\)

Step2: Complete the square

To complete the square for the expression \(x^{2}-4x\), we take half of the coefficient of \(x\), which is \(\frac{-4}{2}=-2\), and then square it: \((-2)^{2} = 4\). We add this value to both sides of the equation:
\(x^{2}-4x + 4=21 + 4\)
The left side can be written as a perfect square: \((x - 2)^{2}=25\)? Wait, no, \(21 + 4=25\)? Wait, no, \(x^{2}-4x+4=(x - 2)^{2}\), and \(21+4 = 25\)? Wait, no, wait the original equation after moving - 31 to the right: \(x^{2}-4x=-10 + 31=21\), then adding 4 to both sides: \((x - 2)^{2}=21 + 4=25\)? Wait, no, that can't be, because if \((x - 2)^{2}=25\), then \(x-2=\pm5\), so \(x=2\pm5\), which gives \(x = 7\) or \(x=-3\). Let's check:

Wait, let's re - do the steps:

Original equation: \(x^{2}-4x-31=-10\)

Step 1: Add 31 to both sides: \(x^{2}-4x=-10 + 31=21\)

Step 2: Take the coefficient of \(x\), which is - 4. Half of - 4 is - 2, square of - 2 is 4. Add 4 to both sides:

\(x^{2}-4x + 4=21 + 4\)

\((x - 2)^{2}=25\)

Step 3: Take the square root of both sides:

\(x - 2=\pm\sqrt{25}=\pm5\)

Step 4: Solve for \(x\):

Case 1: \(x-2 = 5\), then \(x=5 + 2=7\)

Case 2: \(x - 2=-5\), then \(x=-5 + 2=-3\)

Answer:

7, - 3