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Completion requirements
Solve \(-2\left( {x + 3} \right)^2 + \frac{8}{9} = 0\), and verify the solution.
Step 1: Isolate the squared binomial.
\(\begin{align}
-2\left( {x + 3} \right)^2 + \frac{8}{9} &= 0 \\
-2\left( {x + 3} \right)^2 &= - \frac{8}{9} \\
\left( {x + 3} \right)^2 &= \frac{4}{9} \end{align}\)
Step 2: Take the square root of both sides.
\(\begin{align}
\left( {x + 3} \right)^2 &= \frac{4}{9} \\
\sqrt {\left( {x + 3} \right)^2 } &= \pm \sqrt {\frac{4}{9}} \\
x + 3 &= \pm \frac{2}{3} \end{align}\)
Step 3: Solve for \(x\).
\(x = -3 \pm \frac{2}{3}\)
As a result, there are two solutions or roots: \(x = -3 + \frac{2}{3} = -\frac{7}{3}\) and \(x = -3 -\frac{2}{3} = -\frac{{11}}{3}\)
Step 4: Verify.
Verify these solutions by substituting \(-\frac{7}{3}\) and \(-\frac{11}{3}\) for \(x\) in the original equation.
The two sides are equal, so \(x = -\frac{7}{3}\) is a solution.
The two sides are equal, so \(x = -\frac{11}{3}\) has been verified as a solution.
\(\begin{align}
-2\left( {x + 3} \right)^2 + \frac{8}{9} &= 0 \\
-2\left( {x + 3} \right)^2 &= - \frac{8}{9} \\
\left( {x + 3} \right)^2 &= \frac{4}{9} \end{align}\)
Step 2: Take the square root of both sides.
\(\begin{align}
\left( {x + 3} \right)^2 &= \frac{4}{9} \\
\sqrt {\left( {x + 3} \right)^2 } &= \pm \sqrt {\frac{4}{9}} \\
x + 3 &= \pm \frac{2}{3} \end{align}\)
Step 3: Solve for \(x\).
\(x = -3 \pm \frac{2}{3}\)
As a result, there are two solutions or roots: \(x = -3 + \frac{2}{3} = -\frac{7}{3}\) and \(x = -3 -\frac{2}{3} = -\frac{{11}}{3}\)
Step 4: Verify.
Verify these solutions by substituting \(-\frac{7}{3}\) and \(-\frac{11}{3}\) for \(x\) in the original equation.
| Left Side |
Right Side |
|---|---|
| \[\begin{array}{r} - 2\left( {x + 3} \right)^2 + \frac{8}{9} \\ - 2\left( { - \frac{7}{3} + 3} \right)^2 + \frac{8}{9} \\ - 2\left( {\frac{2}{3}} \right)^2 + \frac{8}{9} \\ - 2\left( {\frac{4}{9}} \right) + \frac{8}{9} \\ - \frac{8}{9} + \frac{8}{9} \\ 0 \end{array}\] |
\(0\) |
| LS = RS |
|
The two sides are equal, so \(x = -\frac{7}{3}\) is a solution.
| Left Side |
Right Side |
|---|---|
| \[\begin{array}{r} - 2\left( {x + 3} \right)^2 + \frac{8}{9} \\ - 2\left( { - \frac{{11}}{3} + 3} \right)^2 + \frac{8}{9} \\ - 2\left( { - \frac{2}{3}} \right)^2 + \frac{8}{9} \\ - 2\left( {\frac{4}{9}} \right) + \frac{8}{9} \\ - \frac{8}{9} + \frac{8}{9} \\ 0 \end{array}\] |
\(0\) |
| LS = RS |
|
The two sides are equal, so \(x = -\frac{11}{3}\) has been verified as a solution.
| For further information about Solving Quadratic Equations by Completing the Square see pp. 234 to 240 of Pre-Calculus 11. |