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Completion requirements
Solve \(\sqrt {3 + j} = 2\sqrt j - 3\). Indicate any restrictions on the variable. Round to the nearest hundredth.
Therefore, the restriction on the variable is \(j \ge 0, j \in \rm R\).
\(\begin{align}
\sqrt {3 + j} &= 2\sqrt j - 3 \\
\left( {\sqrt {3 + j} } \right)^2 &= \left( {2\sqrt j - 3} \right)^2 \\
3 + j &= 4j - 12\sqrt j + 9 \\
12\sqrt j &= 3j + 6 \\
4\sqrt j &= j + 2 \\
\left( {4\sqrt j } \right)^2 &= \left( {j + 2} \right)^2 \\
16j &= j^2 + 4j + 4 \\
0 &= j^2 - 12j + 4 \\
\end{align}\)
\(\begin{align}
j &= 6 + 4\sqrt 2 = 11.656... \doteq 11.66 \\
j &= 6 - 4\sqrt 2 = 0.343... \doteq 0.34 \\
\end{align}\)
Verify:
The solution is \(j \doteq 11.66\).
\(\begin{align}
3 + j &\ge 0 \\
j &\ge - 3 \\
\end{align}\)
3 + j &\ge 0 \\
j &\ge - 3 \\
\end{align}\)
\(j \ge 0 \)
Therefore, the restriction on the variable is \(j \ge 0, j \in \rm R\).
\(\begin{align}
\sqrt {3 + j} &= 2\sqrt j - 3 \\
\left( {\sqrt {3 + j} } \right)^2 &= \left( {2\sqrt j - 3} \right)^2 \\
3 + j &= 4j - 12\sqrt j + 9 \\
12\sqrt j &= 3j + 6 \\
4\sqrt j &= j + 2 \\
\left( {4\sqrt j } \right)^2 &= \left( {j + 2} \right)^2 \\
16j &= j^2 + 4j + 4 \\
0 &= j^2 - 12j + 4 \\
\end{align}\)
\[\begin{align}
a &=1, b = -12, c = 4 \\
\\
j &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
j &= \frac{{12 \pm \sqrt {\left( { - 12} \right)^2 - 4\left( 1 \right)\left( 4 \right)} }}{{2\left( 1 \right)}} \\
j &= \frac{{12 \pm \sqrt {144 - 16} }}{2} \\
j &= \frac{{12 \pm \sqrt {128} }}{2} \\
j &= \frac{{12 \pm 8\sqrt 2 }}{2} \\
j &= 6 \pm 4\sqrt 2 \\
\end{align}\]
a &=1, b = -12, c = 4 \\
\\
j &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
j &= \frac{{12 \pm \sqrt {\left( { - 12} \right)^2 - 4\left( 1 \right)\left( 4 \right)} }}{{2\left( 1 \right)}} \\
j &= \frac{{12 \pm \sqrt {144 - 16} }}{2} \\
j &= \frac{{12 \pm \sqrt {128} }}{2} \\
j &= \frac{{12 \pm 8\sqrt 2 }}{2} \\
j &= 6 \pm 4\sqrt 2 \\
\end{align}\]
j &= 6 + 4\sqrt 2 = 11.656... \doteq 11.66 \\
j &= 6 - 4\sqrt 2 = 0.343... \doteq 0.34 \\
\end{align}\)
Verify:
\(j = 6 + 4\sqrt 2 \)
| Left Side |
Right Side |
|---|---|
| \(\begin{array}{r} \sqrt {3 + j} \\ \sqrt {3 + 6 + 4\sqrt 2 } \\ \sqrt {9 + 4\sqrt 2 } \\ 3.828... \\ \end{array}\) |
\(\begin{array} 2\sqrt j - 3 \\ 2\sqrt {6 + 4\sqrt 2 } - 3 \\ 3.828... \\ \end{array}\) |
| LS = RS |
|
\(j = 6 - 4\sqrt 2 \)
This is an extraneous root.
| Left Side |
Right Side |
|---|---|
| \(\begin{array}{r} \sqrt {3 + j} \\ \sqrt {3 + 6 - 4\sqrt 2 } \\ \sqrt {9 - 4\sqrt 2 } \\ 1.828... \\ \end{array}\) |
\(\begin{array} 2\sqrt j - 3 \\ 2\sqrt {6 - 4\sqrt 2 } - 3 \\ - 1.828... \\ \end{array}\) |
| LS \(\ne\) RS |
|
This is an extraneous root.
| For further information about solving radical equations, see pp. 296 – 299 of Pre-Calculus 11. |