# 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) $$\sqrt{x} = 7$$  →  $$(\sqrt{x})^3 =(7)^3$$→  $$x = 243$$      Check: $$\sqrt{243} = 7$$

d) $$\sqrt{x} =- 7$$   →  $$(\sqrt{x})^3 =(-7)^3$$→  $$x = -243$$      Check: $$\sqrt{-243} =- 7$$

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.