Solutions:
a) \(\sqrt {x} = 7\) → \((\sqrt(x))^2 = 7^2\) → \(x = 49\) Check: \(\sqrt{49} = 7\)
b) \(\sqrt{x} = -7 \) → \((\sqrt{x})^2 =(- 7)^2 \) → \(x = 49\) Check: \(\sqrt{49} ≠- 7 \) , Solution is extraneous
c) \(\sqrt[3]{x} = 7\) → \((\sqrt[3]{x})^3 =(7)^3\)→ \(x = 243\) Check: \(\sqrt[3]{243} = 7\)
d) \(\sqrt[3]{x} =- 7\) → \((\sqrt[3]{x})^3 =(-7)^3\)→ \(x = -243\) Check: \(\sqrt[3]{-243} =- 7\)
e) \(\sqrt{x} – 1 = \sqrt{x}\) → \((\sqrt{x} – 1)^2 = (\sqrt{x})^2\)
\(x -2\sqrt{x} + 1 = x\) → \(-2\sqrt{x} = -1\) →\(\sqrt{x} = \Large\frac{1}{2}\)
\(x = \Large\frac{1}{4}\) Check: \(\sqrt{\frac{1}{4}} -1 \neq \sqrt{\frac{1}{4}}\)
f) \(\sqrt{x-15} = \sqrt{x} – 3\) → \((\sqrt{x-15})^2 = (\sqrt{x} – 3 )^2\) → \(x-15 = x – 6\sqrt{x} +9 → 24 = -6\sqrt{x}\)
\(4 =\sqrt{x}\) → \((4)^2=(\sqrt{x})^2\) → \( 16 = x \) Check: \(\sqrt{16-15} = \sqrt{16} – 3\) or \( 1 =4 – 3\)
The reason that b has an extraneous solution and d does not is that a cube of a negative number is negative, but the square of a negative number is positive also e has an extraneous root.