Question

using factoring
$2x^2 + x - 6 = 0$
select the correct response:
$\boldsymbol{ -\frac{2}{3},\\ 2}$
$\boldsymbol{ \frac{3}{2},\\ -2}$
$\boldsymbol{ \frac{2}{3},\\ -2}$
$\boldsymbol{ -\frac{3}{2},\\ 2}$

Explanation:

Step1: Factor the quadratic equation

For the quadratic equation \(2x^{2}+x - 6=0\), we need to find two numbers that multiply to \(2\times(- 6)=-12\) and add up to \(1\). The numbers are \(4\) and \(-3\) since \(4\times(-3)=-12\) and \(4 + (-3)=1\).
Rewrite the middle term using these numbers:
\(2x^{2}+4x-3x - 6 = 0\)
Factor by grouping:
\(2x(x + 2)-3(x + 2)=0\)
Then we get \((2x-3)(x + 2)=0\)

Step2: Solve for \(x\)

Set each factor equal to zero:

• For \(2x-3 = 0\), we have \(2x=3\), so \(x=\frac{3}{2}\)
• For \(x + 2=0\), we have \(x=-2\)

Answer:

\(\frac{3}{2}, -2\) (corresponding to the option \(\frac{3}{2}, -2\))