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Completion requirements
Solve \(3a^2 - 48 = 0\) by factoring. Verify your solution(s).
One side of the equation is \(0\), so factor the quadratic expression. Begin by taking out the GCF.
GCF = \(3\)
\(\begin{align}
3a^2 - 48 &= 0 \\
3\left( {a^2 - 16} \right) &= 0 \end{align}\)
Factor \(a^2 - 16\) as a difference of squares.
\(a^2 - 16 = \left( {a - 4} \right)\left( {a + 4} \right)\)
\(\begin{align}
3\left( {a^2 - 16} \right) = 0 \\
3\left( {a - 4} \right)\left( {a + 4} \right) = 0 \end{align}\)
Determine what values of the variable make each factor equal to zero.
The solutions to \(3a^2 - 48 = 0\) are \(4\) and \(–4\). Verify the solutions by substituting them into the original equation.
GCF = \(3\)
\(\begin{align}
3a^2 - 48 &= 0 \\
3\left( {a^2 - 16} \right) &= 0 \end{align}\)
Factor \(a^2 - 16\) as a difference of squares.
\(a^2 - 16 = \left( {a - 4} \right)\left( {a + 4} \right)\)
\(\begin{align}
3\left( {a^2 - 16} \right) = 0 \\
3\left( {a - 4} \right)\left( {a + 4} \right) = 0 \end{align}\)
Determine what values of the variable make each factor equal to zero.
\(\begin{align}
a - 4 &= 0 \\
a &= 4 \\
\end{align}\)
a - 4 &= 0 \\
a &= 4 \\
\end{align}\)
\(\begin{align}
a + 4 &= 0 \\
a &= -4 \\
\end{align}\)
a + 4 &= 0 \\
a &= -4 \\
\end{align}\)
For \(a = 4\),
The two sides are equal, so \(4\) is a solution.
|
Left Side |
Right Side
|
|---|---|
| \(\begin{array}{r} 3a^2 - 48 \\ 3\left( 4 \right)^2 - 48 \\ 3\left( {16} \right) - 48 \\ 48 - 48 \\ 0 \end{array}\) |
\(0\) |
| LS = RS
|
|
The two sides are equal, so \(4\) is a solution.
For \(a = -4\),
The two sides are equal, so \(–4\) is a solution.
|
Left Side
|
Right Side
|
|---|---|
| \(\begin{array}{r} 3a^2 - 48 \\ 3\left( { -4} \right)^2 - 48 \\ 3\left( {16} \right) - 48 \\ 48 - 48 \\ 0 \end{array}\) |
\(0\) |
|
LS = RS
|
|
The two sides are equal, so \(–4\) is a solution.
| For further information about Factoring Quadratic Equations see pp. 218 to 229 of Pre-Calculus 11. |