Question

Which are the endpoints of the solution intervals for the quadratic inequality x^2+16x>16?A. X=-8 and X=2B. X=-8 and Y=-2C. X=8 and Y=2D. X=8 and X=2

Answer 1

We are given the following quadratic inequality

$x^2+6x>16$

Let us first convert it into quadratic form and then factorize the inequality

$x^2_{}+6x-16>0$

Now we need two numbers such that their sum is 6 and their product is -16

How about 8 and -2?

Sum = 8 - 2 = 6

Product = 8*-2 = 16

$\begin{gathered} x^2_{}+6x-16>0 \\ (x+8)(x-2)>0_{} \end{gathered}$

Now there are a few cases possible,

$\begin{gathered} (x+8)>0\quad and\quad (x-2)>0_{} \\ x>-8\quad and\quad x>2_{} \end{gathered}$

Since x > 2 meets the requirement of x > -8 so we take x > 2 from here.

$\begin{gathered} (x+8)<0\quad and\quad (x-2)<0 \\ x<-8\quad and\quad x<2 \end{gathered}$

Since x < -8 meets the requirement of x < 2 so we take x < -8 from here.

Therefore, the solution of the quadratic inequality is

$x<-8\quad or\quad x>2$

The endpoints are x = -8 and x = 2 (option A is correct)