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

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)

anonymous · Answer

$$9x ^{2}+x \le8$$

Subtract 8,

$$9x ^{2}+x -8 \le 0$$
Factored:
(x+1)(9x-8)

So to write this:
$$-1\le x \le8/9$$

anonymous · Answer

$$9x^2+x \le8$$$$9x^2+x -8\le0$$
$$Δ=1^2-4\times9(-8)=289$$$$x_{1,2}=\frac{-1\pm \sqrt{289}}{2\times9}$$$$x_{1,2}=\frac{-1\pm 17}{18}$$$$x_1=\frac{-1+17}{18}=\frac{8}{9}$$$$x_2=\frac{-1-17}{18}=-1$$|dw:1315752070730:dw|

So, finally, we get$$x \in[-1,\frac{8}{9}]$$