Example 2
Completion requirements
| Example 2 |
Solve \(12m^2 - 5m = 3\) by factoring, and then verify the solution.
Step 1:
Rearrange the equation so one side of the equal sign is zero.
\(\begin{align}
12m^2 - 5m &= 3 \\
12m^2 - 5m - 3 &= 0 \\
\end{align}\)
Step 2: Factor the quadratic expression.
\(12m^2 - 5m - 3 = 0\)
Product of \(12(–3) = –36\), sum of \(–5 \)
The two numbers are \(–9\) and \(4\).
Decompose the middle term, group, and factor.
\(\begin{align}
12m^2 - 5m - 3 &= 0 \\
12m^2 - 9m + 4m - 3 &= 0 \\
3m\left( {4m - 3} \right) + \left( {4m - 3} \right) &= 0 \\
\left( {4m - 3} \right)\left( {3m + 1} \right) &= 0 \\
\end{align}\)
Step 3: Determine what value of the variable makes each factor equal to zero.
The solutions are \(\frac{3}{4}\) and \(-\frac{1}{3}\).
Step 4: Verify the solutions by substituting them into the original equation.
\(\begin{align}
12m^2 - 5m &= 3 \\
12m^2 - 5m - 3 &= 0 \\
\end{align}\)
Step 2: Factor the quadratic expression.
\(12m^2 - 5m - 3 = 0\)
Product of \(12(–3) = –36\), sum of \(–5 \)
The two numbers are \(–9\) and \(4\).
Decompose the middle term, group, and factor.
\(\begin{align}
12m^2 - 5m - 3 &= 0 \\
12m^2 - 9m + 4m - 3 &= 0 \\
3m\left( {4m - 3} \right) + \left( {4m - 3} \right) &= 0 \\
\left( {4m - 3} \right)\left( {3m + 1} \right) &= 0 \\
\end{align}\)
Step 3: Determine what value of the variable makes each factor equal to zero.
\(\begin{align}
4m - 3 &= 0 \\
4m &= 3 \\
m &= \frac{3}{4} \end{align}\)
4m - 3 &= 0 \\
4m &= 3 \\
m &= \frac{3}{4} \end{align}\)
\(\begin{align}
3m + 1 &= 0 \\
3m &= -1 \\
m &= -\frac{1}{3}
\end{align}\)
3m + 1 &= 0 \\
3m &= -1 \\
m &= -\frac{1}{3}
\end{align}\)
The solutions are \(\frac{3}{4}\) and \(-\frac{1}{3}\).
Step 4: Verify the solutions by substituting them into the original equation.
For \(m = \frac{3}{4}\),
The two sides are equal, so \(\frac{3}{4}\) is a solution.
|
Left Side
|
Right Side
|
|---|---|
|
\[\begin{array}{r}
12m^2 - 5m \\ 12\left( {\frac{3}{4}} \right)^2 - 5\left( {\frac{3}{4}} \right) \\ 12\left( {\frac{9}{{16}}} \right) - \frac{{15}}{4} \\ \frac{{27}}{4} - \frac{{15}}{4} \\ \frac{{12}}{4} \\ 3 \end{array}\] |
\(3\) |
| LS = RS
|
|
The two sides are equal, so \(\frac{3}{4}\) is a solution.
For \(m = -\frac{1}{3}\),
The two sides are equal, so \(-\frac{1}{3}\) is a solution.
|
Left Side
|
Right Side
|
|---|---|
|
\[\begin{array}{r}
12m^2 - 5m \\ 12\left( {-\frac{1}{3}} \right)^2 - 5\left( { - \frac{1}{3}} \right) \\ 12\left( {\frac{1}{9}} \right) + \frac{5}{3} \\ \frac{4}{3} + \frac{5}{3} \\ \frac{9}{3} \\ 3 \end{array}\] |
\(3
\) |
| LS = RS
|
|
The two sides are equal, so \(-\frac{1}{3}\) is a solution.