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Completion requirements
Solve the inequality \(-x^2 + 8x - 12 \lt 0\). Plot the solution set on a number line.
Determine the zeros of the corresponding function.
\(\begin{align}
- x^2 + 8x - 12 &= 0 \\
- 1\left( {x - 6} \right)\left( {x - 2} \right) &= 0
\end{align}\)
The intervals that may be part of the solution set are:
The zeros are not part of the solution set because the inequality is strict. Use test points to determine which intervals are part of the solution set.
Test \(x \lt 2 \), using \(x = 0\).
The test point \(0\) satisfies the original inequality, so \(x \lt 2\) is part of the solution set.
Test \(2 \lt x \lt 6 \), using \(x = 3\).
The test point \(3\) does not satisfy the original inequality, so \(2 \lt x \lt 6\) is not part of the solution set.
Test \(x \gt 6 \), using \(x = 8\).
The test point \(8\) satisfies the original inequality, so \(x \gt 6\) is part of the solution set.
Combine the appropriate solution intervals to show the entire solution set.
{\(x | x \lt 2 \thinspace {\rm{or}} \thinspace x \gt 6, \thinspace x \in \rm{R}\)}
Plot the solution set on a number line.
\(\begin{align}
- x^2 + 8x - 12 &= 0 \\
- 1\left( {x - 6} \right)\left( {x - 2} \right) &= 0
\end{align}\)
\(\begin{align}
x - 6 &= 0 \\
x &= 6
\end{align}\)
x - 6 &= 0 \\
x &= 6
\end{align}\)
\(\begin{align}
x - 2 &= 0 \\
x &= 2
\end{align}\)
x - 2 &= 0 \\
x &= 2
\end{align}\)
The intervals that may be part of the solution set are:
- \(x \lt 2\)
- \(2 \lt x \lt 6\)
- \(x \gt 6\)
The zeros are not part of the solution set because the inequality is strict. Use test points to determine which intervals are part of the solution set.
Test \(x \lt 2 \), using \(x = 0\).
| Left Side | Right Side |
|---|---|
|
\(\begin{array}{r} - x^2 + 8x - 12 \\ - \left( 0 \right)^2 + 8\left( 0 \right) - 12 \\ - 12 \end{array}\) |
\(0\) |
| LS \(\lt\) RS \(\hspace{30pt}\) | |
The test point \(0\) satisfies the original inequality, so \(x \lt 2\) is part of the solution set.
Test \(2 \lt x \lt 6 \), using \(x = 3\).
| Left Side | Right Side |
|---|---|
|
\(\begin{array}{r} - x^2 + 8x - 12 \\ - \left( 3 \right)^2 + 8\left( 3 \right) - 12 \\ 3 \end{array}\) |
\(0\) |
| LS \(\gt\) RS \(\hspace{30pt}\) | |
The test point \(3\) does not satisfy the original inequality, so \(2 \lt x \lt 6\) is not part of the solution set.
Test \(x \gt 6 \), using \(x = 8\).
| Left Side | Right Side |
|---|---|
|
\(\begin{array}{r} - x^2 + 8x - 12 \\ - \left( 8 \right)^2 + 8\left( 8 \right) - 12 \\ - 12 \end{array}\) |
\(0\) |
| LS \(\lt\) RS \(\hspace{30pt}\) | |
The test point \(8\) satisfies the original inequality, so \(x \gt 6\) is part of the solution set.
Combine the appropriate solution intervals to show the entire solution set.
{\(x | x \lt 2 \thinspace {\rm{or}} \thinspace x \gt 6, \thinspace x \in \rm{R}\)}
Plot the solution set on a number line.