POLQ 2| Lesson 3| Solutions (Solving Radical Equations)

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.


Return to Practice (Solving Radical Equations)