Solve for x if 2(5x + 2)2 = 48.

Question

x = StartFraction negative 2 + 2 StartRoot 24 EndRoot Over 5 EndFraction and x = StartFraction negative 2 minus 2 StartRoot 24 EndRoot Over 5 EndFraction
x = StartFraction negative 2 + 2 StartRoot 24 EndRoot Over 2 EndFraction and x = StartFraction negative 2 minus 2 StartRoot 24 EndRoot Over 2 EndFraction
x = StartFraction negative 2 + 2 StartRoot 6 EndRoot Over 5 EndFraction and x = StartFraction negative 2 minus 2 StartRoot 6 EndRoot Over 5 EndFraction
x = StartFraction negative 2 + 2 StartRoot 6 EndRoot Over 2 EndFraction and x = StartFraction negative 2 minus 2 StartRoot 6 EndRoot Over 2 EndFraction

Answer 1

To solve the equation \( 2(5x + 2)^2 = 48 \), let's start by isolating \( (5x + 2)^2 \):

1. Divide both sides by 2: \[ (5x + 2)^2 = \frac{48}{2} = 24 \]

2. Take the square root of both sides: \[ 5x + 2 = \pm \sqrt{24} \] We can simplify \( \sqrt{24} \): \[ \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6} \] So we have: \[ 5x + 2 = 2\sqrt{6} \quad \text{and} \quad 5x + 2 = -2\sqrt{6} \]

3. Now, solve for \( x \) in each case.

Case 1: \[ 5x + 2 = 2\sqrt{6} \] Subtract 2 from both sides: \[ 5x = 2\sqrt{6} - 2 \] Divide by 5: \[ x = \frac{2\sqrt{6} - 2}{5} \]

Case 2: \[ 5x + 2 = -2\sqrt{6} \] Subtract 2 from both sides: \[ 5x = -2\sqrt{6} - 2 \] Divide by 5: \[ x = \frac{-2\sqrt{6} - 2}{5} \]

Thus, the solutions for \( x \) are: \[ x = \frac{2\sqrt{6} - 2}{5} \quad \text{and} \quad x = \frac{-2\sqrt{6} - 2}{5} \]

These simplifications can also be written as: \[ x = \frac{-2 + 2\sqrt{6}}{5} \quad \text{and} \quad x = \frac{-2 - 2\sqrt{6}}{5} \]

Therefore, the answers match with: \[ x = \frac{-2 + 2\sqrt{6}}{5} \text{ and } x = \frac{-2 - 2\sqrt{6}}{5} \]

Hence, the correct answer corresponds to: \[ x = \frac{-2 + 2\sqrt{6}}{5} \text{ and } x = \frac{-2 - 2\sqrt{6}}{5} \]