C)[A.REI.A.2] [10 PTS] DIRECTIONS: Solve each equation and identify any extrancous solutions. 1) \( \sqrt{x+1}=\sqrt{7 x+15} \) 2) \( (2 x)^{\frac{3}{4}}+2=10 \) 3) \( \sqrt[3]{x}-10=-7 \) 4) \( -2 \sqrt{24 x}-15=17 \)

To solve the equations, let's break them down one by one: 1) **Equation**: \( \sqrt{x+1}=\sqrt{7 x+15} \) **Solution**: Square both sides: \( x + 1 = 7x + 15 \) Rearranging gives: \( 1 - 15 = 7x - x \) So, \( -14 = 6x \) or \( x = -\frac{7}{3} \). **Check**: When substituted back, each side equals \( \sqrt{-\frac{7}{3}+1} \) which results in a complex number, hence no real solutions. 2) **Equation**: \( (2 x)^{\frac{3}{4}} + 2 = 10 \) **Solution**: Subtract 2 from both sides: \( (2x)^{\frac{3}{4}} = 8 \) Raise both sides to \( \frac{4}{3} \): \( 2x = 8^{\frac{4}{3}} = 16 \) Thus, \( 2x = 16 \) or \( x = 8 \). **Check**: \( (2 \cdot 8)^{\frac{3}{4}} + 2 = 10 \) holds true. 3) **Equation**: \( \sqrt[3]{x}-10=-7 \) **Solution**: Add 10 to both sides: \( \sqrt[3]{x} = 3 \) Now cube both sides: \( x = 3^3 = 27 \). **Check**: \( \sqrt[3]{27}-10 = -7 \) confirms the solution is valid. 4) **Equation**: \( -2 \sqrt{24x}-15=17 \) **Solution**: Add 15 to both sides: \( -2\sqrt{24x} = 32 \) Divide by -2: \( \sqrt{24x} = -16 \), which is impossible in real numbers. Thus, there are no real solutions. In conclusion, the extraneous solutions encountered show how careful handling is necessary when dealing with square roots and cube roots. Always check your work!