${cases} x^2+x-6 0 \ -2x^2+3x+2>0 {cases}$See answer

Daftar Isi

cases x2 x 6 0 2×2 3x 20 cases

Question

$\begin{cases} x^2+x-6 \ge 0 \ -2x^2+3x+2>0 \end{cases}$

Basic Answer

Step 1: Solve the first inequality $x^2+x-6\ge 0$

1. Factor the quadratic expression:
   $$x^2+x-6=(x+3)(x-2)$$

2. Set up the inequality:
   $$(x+3)(x-2)\ge 0$$

3. Determine the critical points by setting each factor to zero:
   $$x+3=0\Rightarrow x=-3$$
   $$x-2=0\Rightarrow x=2$$

4. Use a sign chart to determine where the product is non-negative:

   • For $x<-3$, both factors are negative, so the product is positive.
   • For $-3<x<2$, one factor is negative and the other is positive, so the product is negative.
   • For $x>2$, both factors are positive, so the product is positive.

5. Combine the intervals:
   $$x\in (-\infty ,-3]\cup [2,\infty )$$

Step 2: Solve the second inequality $-2x^2+3x+2>0$

1. Factor the quadratic expression:
   $$-2x^2+3x+2=-(2x^2-3x-2)=-(2x+1)(x-2)$$

2. Set up the inequality:
   $$-(2x+1)(x-2)>0$$

3. Determine the critical points by setting each factor to zero:
   $$2x+1=0\Rightarrow x=-\frac{1}{2}$$
   $$x-2=0\Rightarrow x=2$$

4. Use a sign chart to determine where the product is positive:

   • For $x<-\frac{1}{2}$, both factors are negative, so the product is negative.
   • For $-\frac{1}{2}<x<2$, one factor is negative and the other is positive, so the product is positive.
   • For $x>2$, both factors are positive, so the product is negative.

5. Combine the intervals:
   $$x\in \left(-\frac{1}{2},2\right)$$

Step 3: Find the intersection of the solutions

1. The solution to the first inequality is:
   $$x\in (-\infty ,-3]\cup [2,\infty )$$

2. The solution to the second inequality is:
   $$x\in \left(-\frac{1}{2},2\right)$$

3. The intersection of these intervals is:
   $$x\in \left(-\frac{1}{2},2\right)$$

Final Answer

$$x\in \left(-\frac{1}{2},2\right)$$