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

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:  $$\sqrt{49} \neq- 7$$ , $$-7$$ is an extraneous solution for $$x^2=49$$. The solution for $$x$$ is no solution.

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) $$\sqrt{x}-1 = \sqrt{x}$$  → $$(\sqrt{x}-1)^2 = (\sqrt{x})^2$$

$$x-2\sqrt{x}+1 = x$$ →  $$-2\sqrt{x} = -1$$ →$$\sqrt{x} = \dfrac{1}{2}$$

$$x = \dfrac{1}{4}$$

Check:

\begin{align}\dfrac14-2\sqrt{\dfrac14}+1 &= \dfrac14 \\\\ \dfrac14-2\bigg(\dfrac12\bigg)+1&=\dfrac14\\\\ \dfrac14 -1+1 &= \dfrac14 \\\\ \dfrac14&=\dfrac14\end{align}

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.