Question

cc.2.2.hs.d.8
solve
$(x - 5)^2 = 49$
$\bigcirc$ $x = -2$ and $x = 12$
$\bigcirc$ $x = 2$ and $x = 7$
$\bigcirc$ $x = 54$
$\bigcirc$ $x = -3$ and $x = 12$

Explanation:

Step1: Take square root of both sides

To solve the equation \((x - 5)^2 = 49\), we take the square root of both sides. Remember that when we take the square root of a number, we get both a positive and a negative solution. So we have:
\[
x - 5=\pm\sqrt{49}
\]
Since \(\sqrt{49} = 7\), this simplifies to:
\[
x - 5=\pm7
\]

Step2: Solve for \(x\) in both cases

Case 1: When \(x - 5 = 7\)

Add 5 to both sides of the equation:
\[
x=7 + 5
\]
\[
x = 12
\]

Case 2: When \(x - 5=-7\)

Add 5 to both sides of the equation:
\[
x=-7 + 5
\]
\[
x=-2
\]

Answer:

\(x=-2\) and \(x = 12\) (corresponding to the first option: \(x=-2\) and \(x = 12\))