Question

solve the quadratic inequality. state the solution as an inequality. 9x² < 64
-8/3 ≤ x ≤ 8/3
x < -8/3 or x > 8/3
x ≤ -8/3 or x ≥ 8/3
-8/3 < x < 8/3

Explanation:

Step1: Rearrange the inequality

$9x^{2}-64<0$

Step2: Factor the left - hand side

Using the difference of squares formula $a^{2}-b^{2}=(a + b)(a - b)$, where $a = 3x$ and $b = 8$, we get $(3x + 8)(3x-8)<0$.

Step3: Find the roots

Set each factor equal to zero: $3x+8 = 0$ gives $x=-\frac{8}{3}$, and $3x - 8=0$ gives $x=\frac{8}{3}$.

Step4: Test intervals

We have three intervals to test: $x<-\frac{8}{3}$, $-\frac{8}{3}\frac{8}{3}$.
For $x<-\frac{8}{3}$, let $x=-3$, then $(3\times(-3)+8)(3\times(-3)-8)=(-1)(-17)=17>0$.
For $-\frac{8}{3}For $x>\frac{8}{3}$, let $x = 3$, then $(3\times3+8)(3\times3 - 8)=(17)(1)=17>0$.

Answer:

$-\frac{8}{3}