Question

solve for x.
$-2x^{2} + 5x + 11 > -5x - 1$
write a compound inequality like $1 < x < 3$ or like $x < 1$ or $x > 3$.

Explanation:

Step1: Move all terms to left side

First, we add \(5x\) and \(1\) to both sides of the inequality \(-2x^{2}+5x + 11>-5x - 1\) to get a quadratic inequality in standard form.
\(-2x^{2}+5x + 11+5x + 1>0\)
Simplify the left - hand side: \(-2x^{2}+(5x + 5x)+(11 + 1)>0\), which is \(-2x^{2}+10x + 12>0\)
We can multiply both sides by \(- 1\) to make the coefficient of \(x^{2}\) positive. Remember that when we multiply or divide an inequality by a negative number, the direction of the inequality sign changes. So we get \(2x^{2}-10x - 12<0\)
Then divide the entire inequality by \(2\) to simplify: \(x^{2}-5x - 6<0\)

Step2: Factor the quadratic expression

Factor the quadratic \(x^{2}-5x - 6\). We need two numbers that multiply to \(-6\) and add up to \(-5\). The numbers are \(-6\) and \(1\).
So \(x^{2}-5x - 6=(x - 6)(x+1)\)
The inequality becomes \((x - 6)(x + 1)<0\)

Step3: Find the critical points

Set each factor equal to zero and solve for \(x\):
For \(x-6 = 0\), we have \(x = 6\)
For \(x + 1=0\), we have \(x=-1\)

Step4: Analyze the sign of the product

The critical points \(x=-1\) and \(x = 6\) divide the number line into three intervals:

• Interval 1: \(x<-1\). Let's test \(x=-2\) in \((x - 6)(x + 1)\). \((-2-6)(-2 + 1)=(-8)\times(-1)=8>0\)
• Interval 2: \(-1
• Interval 3: \(x>6\). Let's test \(x = 7\) in \((x - 6)(x + 1)\). \((7 - 6)(7 + 1)=(1)\times(8)=8>0\)

Since we want \((x - 6)(x + 1)<0\), the solution is the interval where the product is negative, which is \(-1

Answer:

\(-1