Question

which is the solution of the quadratic equation ((4y - 3)^2 = 72)?
(y = \frac{9sqrt{2}}{4}) and (y = \frac{-3sqrt{2}}{4})
(y = \frac{3 + 6sqrt{2}}{4}) and (y = \frac{-3 - 6sqrt{2}}{4})
(y = \frac{3 + 6sqrt{2}}{4}) and (y = \frac{3 - 6sqrt{2}}{4})

Explanation:

Step1: Take square root of both sides

Given the equation \((4y - 3)^2 = 72\), we take the square root of both sides. Remember that if \(x^2=a\), then \(x = \pm\sqrt{a}\). So we have:
\(4y - 3=\pm\sqrt{72}\)
Simplify \(\sqrt{72}\), we know that \(72 = 36\times2\), so \(\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}\). So the equation becomes:
\(4y - 3=\pm6\sqrt{2}\)

Step2: Solve for y when taking the positive square root

First, consider the case when we take the positive square root, i.e., \(4y - 3 = 6\sqrt{2}\).
Add 3 to both sides:
\(4y=3 + 6\sqrt{2}\)
Divide both sides by 4:
\(y=\frac{3 + 6\sqrt{2}}{4}\)

Step3: Solve for y when taking the negative square root

Next, consider the case when we take the negative square root, i.e., \(4y - 3=-6\sqrt{2}\).
Add 3 to both sides:
\(4y=3-6\sqrt{2}\)
Divide both sides by 4:
\(y=\frac{3 - 6\sqrt{2}}{4}\)

Answer:

\(y=\frac{3 + 6\sqrt{2}}{4}\) and \(y=\frac{3 - 6\sqrt{2}}{4}\) (the third option)