Example 1
Completion requirements
| Example 1 |
Solve \(\left( {x - 1} \right)^2 - 36 = 0\), and verify the solution.
Step 1: Isolate the squared binomial.
Note that in order to get a positive number before you take the square root, \(a\) and \(q\) must be opposite signs!
Step 2: Take the square root of both sides.
\(\begin{align}
\sqrt {\left( {x - 1} \right)^2 } &= \pm \sqrt {36} \\
x - 1 &= \pm 6 \\
\end{align}\)
Step 3: Solve for \(x\).
\(x = 1 \pm 6\)
As a result, there are two solutions or roots: \(x = 1 + 6 = 7\) and \(x = 1 - 6 = -5\)
Step 4: Verify.
Verify these solutions by substituting \(–5\) and \(7\) for \(x\) in the original equation.
The two sides are equal, so \(x = –5\) is a solution.
The two sides are equal, so \(x = 7\) is a solution.
\(\begin{align}
\left( {x - 1} \right)^2 - 36 &= 0 \\
\left( {x - 1} \right)^2 &= 36 \\
\end{align}\)
\left( {x - 1} \right)^2 - 36 &= 0 \\
\left( {x - 1} \right)^2 &= 36 \\
\end{align}\)
Note that in order to get a positive number before you take the square root, \(a\) and \(q\) must be opposite signs!
Step 2: Take the square root of both sides.
\(\begin{align}
\sqrt {\left( {x - 1} \right)^2 } &= \pm \sqrt {36} \\
x - 1 &= \pm 6 \\
\end{align}\)
Step 3: Solve for \(x\).
\(x = 1 \pm 6\)
As a result, there are two solutions or roots: \(x = 1 + 6 = 7\) and \(x = 1 - 6 = -5\)
Step 4: Verify.
Verify these solutions by substituting \(–5\) and \(7\) for \(x\) in the original equation.
| Left Side |
Right Side |
|---|---|
| \(\begin{array}{r} \left( {x - 1} \right)^2 - 36 \\ \left( { - 5 - 1} \right)^2 - 36 \\ \left( { - 6} \right)^2 - 36 \\ 36 - 36 \\ 0 \end{array}\) |
\(0\) |
| LS = RS |
|
The two sides are equal, so \(x = –5\) is a solution.
| Left Side |
Right Side |
|---|---|
| \(\begin{array}{r} \left( {x - 1} \right)^2 - 36 \\ \left( {7 - 1} \right)^2 - 36 \\ \left( 6 \right)^2 - 36 \\ 36 - 36 \\ 0 \end{array}\) |
\(0\) |
| LS = RS |
|
The two sides are equal, so \(x = 7\) is a solution.