Question

how to solve the inequality in terms of intervals and illustrate the solution set on the real number line.
9x2( nine X square)+x< and = 8

( I could not write square in this site)

Answer 1

\[9x ^{2}+x \le8\]

Subtract 8,

\[9x ^{2}+x -8 \le 0\]
Factored:
(x+1)(9x-8)

So to write this:
\[-1\le x \le8/9\]

Answer 2

\[9x^2+x \le8\]\[9x^2+x -8\le0\]
\[Δ=1^2-4\times9(-8)=289\]\[x_{1,2}=\frac{-1\pm \sqrt{289}}{2\times9}\]\[x_{1,2}=\frac{-1\pm 17}{18}\]\[x_1=\frac{-1+17}{18}=\frac{8}{9}\]\[x_2=\frac{-1-17}{18}=-1\]

So, finally, we get\[x \in[-1,\frac{8}{9}]\]