Question

Find the zeros of each functions by using a table. F(x)=2x^2+2x-24

Answer 1

We have to find the zeros of the function:

$f(x)=2x^2+2x-24$

We can solve this with the quadratic equation, but first we can divide all the coefficients by a factor of 2:

$\begin{gathered} 2x^2+2x-24=0 \\ 2(x^2+x-12)=0 \\ x^2+x-12=0 \end{gathered}$

Then, we apply the quadratic equation:

$\begin{gathered} x=\frac{-b}{2a}\pm\frac{\sqrt[]{b^2-4ac}}{2a} \\ \\ x=-\frac{1}{2\cdot1}\pm\frac{\sqrt[]{1-4\cdot1\cdot(-12)}}{2\cdot1} \\ \\ x=-\frac{1}{2}\pm\frac{\sqrt[]{1+48}}{2}=-\frac{1}{2}\pm\frac{\sqrt[]{49}}{2}=-\frac{1}{2}\pm\frac{7}{2} \\ x_1=-\frac{1}{2}+\frac{7}{2}=\frac{6}{2}=3 \\ x_2=-\frac{1}{2}-\frac{7}{2}=-\frac{8}{2}=-4 \end{gathered}$

The zeros of the function are x1=3 and x2=-4.