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

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