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.


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