The sum of the values of \(x\) that cancel the expression \(2x^2 - 4x - 16\) is:
-
\(-4\)
-
\(-2\)
-
\(2\)
-
\(4\)
-
\(8\)
Try to solve it before checking the answer.
\(2\)
We are being asked for the sum of the roots of the equation: \( 2x^2 – 4x – 16 = 0 \). Well,
\begin{aligned}
&2 x^2 – 4x – 16 = 0
\\[.5em]
&\hspace{3em}\Leftrightarrow
2 \left( x^2 – 2x – 8 \right) = 0
\\[.5em]
&\hspace{3em}\Leftrightarrow x^2 – 2x – 8 = 0
\\[.5em]
&\hspace{3em}\Leftrightarrow
(x – 4) (x + 2) = 0
\\[.5em]
&\hspace{3em}\Leftrightarrow
x – 4 = 0 \quad \text{ó} \quad x + 2= 0
\\[.5em]
&\hspace{3em}\Leftrightarrow
x = 4 \quad \text{ó} \quad x = -2
\end{aligned}
\]
\begin{aligned}
2 x^2 – 4x – 16 = 0
&\Leftrightarrow
2 \left( x^2 – 2x – 8 \right) = 0
\\[.5em]
&\Leftrightarrow x^2 – 2x – 8 = 0
\\[.5em]
&\Leftrightarrow
(x – 4) (x + 2) = 0
\\[.5em]
&\Leftrightarrow
x – 4 = 0 \quad \text{ó} \quad x + 2= 0
\\[.5em]
&\Leftrightarrow
x = 4 \quad \text{ó} \quad x = -2
\end{aligned}
\]
The sum of these roots is:
4 + (-2) = \boldsymbol{2}
\]
Remark: We can easily obtain this result just keeping in mind that, in the equation \(x^2 + bx + c = 0\), the sum of its roots is \(-b\) and its product is \(c\).