Question

determine the solution set of the quadratic inequality: 2x^2 + x + 8 > 11. your answer should be in interval notation.
(-∞, - 3/2) ∪ (1, ∞)
(- 3/2, 1)
(-∞, - 1) ∪ (3/2, ∞)
(-1, 3/2)

Explanation:

Step1: Rearrange the inequality

Subtract 11 from both sides: $2x^{2}+x + 8-11>0$, which simplifies to $2x^{2}+x - 3>0$.

Step2: Factor the quadratic expression

Factor $2x^{2}+x - 3$ as $(2x + 3)(x - 1)>0$.

Step3: Find the roots

Set each factor equal to zero: $2x+3 = 0$ gives $x=-\frac{3}{2}$, and $x - 1=0$ gives $x = 1$.

Step4: Test intervals

Test the intervals $(-\infty,-\frac{3}{2})$, $(-\frac{3}{2},1)$ and $(1,\infty)$.
For $x<-\frac{3}{2}$, let $x=-2$. Then $(2(-2)+3)(-2 - 1)=(-1)\times(-3)=3>0$.
For $-\frac{3}{2}For $x>1$, let $x = 2$. Then $(2\times2+3)(2 - 1)=(7)\times(1)=7>0$.

Answer:

$(-\infty,-\frac{3}{2})\cup(1,\infty)$