Exercise 44
Solving the equation: \( \sqrt{ 5 (x + 33) } = x + 3 \), yields the following solution set:
-
\( \{ 12, \, 13 \} \)
-
\( \{ -12, \, 13 \} \)
-
\( \{ 12, -13 \} \)
-
\( \{ -12, -13 \} \)
-
\( \{ 12 \} \)
Try to solve it before checking the answer.
-
\( \{ 12, -13 \} \)
\[
\begin{aligned}
&\sqrt{ 5 (x + 33) } = x + 3
\\[.5em]
&\hspace{4em}\Leftrightarrow
5 (x + 33) = (x + 3)^3
\\
&\hspace{6.5em}
\text{(squaring)}
\\[.5em]
&\hspace{4em}\Leftrightarrow
5x + 165 = x^2 + 6x + 9
\\[.5em]
&\hspace{4em}\Leftrightarrow
x^2 + x – 156 = 0
\\[.5em]
&\hspace{4em}\Leftrightarrow
(x + 13)(x – 12) = 0
\\[.5em]
&\hspace{4em}\Leftrightarrow
x + 13 = 0 \quad \text{ or } \quad
x – 12 = 0
\\[.5em]
&\hspace{4em}\Leftrightarrow
\boldsymbol{x = -13} \quad \text{ or } \quad \boldsymbol{ x = 12 }
\end{aligned}
\]
\[
\begin{aligned}
\sqrt{ 5 (x + 33) } = x + 3
&\Leftrightarrow
5 (x + 33) = (x + 3)^3
\quad
\text{(squaring)}
\\[.5em]
&\Leftrightarrow
5x + 165 = x^2 + 6x + 9
\\[.5em]
&\Leftrightarrow
x^2 + x – 156 = 0
\\[.5em]
&\Leftrightarrow
(x + 13)(x – 12) = 0
\\[.5em]
&\Leftrightarrow
x + 13 = 0 \quad \text{ or } \quad
x – 12 = 0
\\[.5em]
&\Leftrightarrow
\boldsymbol{x = -13} \quad \text{ or } \quad \boldsymbol{ x = 12 }
\end{aligned}
\]