Question

what are the possible values of x if $(4x - 5)^2 = 49$? check all that apply. $-\frac{4}{5}$ $-\frac{1}{2}$ 3 5 7

Explanation:

Step1: Take square root of both sides

To solve \((4x - 5)^2 = 49\), we take the square root of both sides. Remember that if \(y^2 = a\) (where \(a\geq0\)), then \(y=\pm\sqrt{a}\). So applying this to our equation, we get:
\(4x - 5=\pm\sqrt{49}\)
Since \(\sqrt{49} = 7\), this simplifies to:
\(4x - 5=\pm7\)

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

Case 1: \(4x - 5 = 7\)

Add 5 to both sides of the equation:
\(4x-5 + 5=7 + 5\)
\(4x=12\)
Divide both sides by 4:
\(x=\frac{12}{4}=3\)

Case 2: \(4x - 5=-7\)

Add 5 to both sides of the equation:
\(4x-5 + 5=-7 + 5\)
\(4x=-2\)
Divide both sides by 4:
\(x=\frac{-2}{4}=-\frac{1}{2}\)

Answer:

\(-\frac{1}{2}\), \(3\) (corresponding to the options: \(-\frac{1}{2}\) and \(3\))