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.
Return to Practice (Solving Radical Equations)