Question

what are the solutions to the quadratic equation $(5y + 6)^2 = 24$?

$\bigcirc$ $y = \frac{-6 + 2\sqrt{6}}{5}$ and $y = \frac{-6 - 2\sqrt{6}}{5}$

$\bigcirc$ $y = \frac{-6 + 2\sqrt{6}}{5}$ and $y = \frac{6 - 2\sqrt{6}}{5}$

$\bigcirc$ $y = \frac{-4\sqrt{6}}{5}$ and $y = \frac{-8\sqrt{6}}{5}$

$\bigcirc$ $y = \frac{4\sqrt{6}}{5}$ and $y = \frac{8\sqrt{6}}{5}$

Explanation:

Step1: Take square root of both sides

To solve \((5y + 6)^2 = 24\), we first take the square root of both sides. Remember that when we take the square root of a square, we get both a positive and a negative solution. So we have:

$$ 5y + 6=\pm\sqrt{24} $$

Simplify \(\sqrt{24}\). We know that \(24 = 4\times6\), so \(\sqrt{24}=\sqrt{4\times6} = 2\sqrt{6}\) (since \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) for \(a\geq0,b\geq0\) and \(\sqrt{4} = 2\)). So now our equation becomes:

$$ 5y+6=\pm2\sqrt{6} $$

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

Case 1: When \(5y + 6=2\sqrt{6}\)

Subtract 6 from both sides:

$$ 5y=2\sqrt{6}-6 $$

Then divide both sides by 5:

$$ y=\frac{2\sqrt{6}-6}{5}=\frac{-6 + 2\sqrt{6}}{5} $$

Case 2: When \(5y + 6=- 2\sqrt{6}\)

Subtract 6 from both sides:

$$ 5y=-2\sqrt{6}-6 $$

Then divide both sides by 5:

$$ y=\frac{-2\sqrt{6}-6}{5}=\frac{-6-2\sqrt{6}}{5} $$

Answer:

\(y=\frac{-6 + 2\sqrt{6}}{5}\) and \(y=\frac{-6-2\sqrt{6}}{5}\) (the first option)