Solve the quadratic inequality 3x2+3x−6≤0 by factoring.(1 point)

Question

Responses

−2≤x≤1
negative 2 less-than-or-equal-to x less-than-or-equal-to 1

−1<x<2
negative 1 less than x less than 2

x≤−2 or x≥3
x less-than-or-equal-to negative 2 or x greater-than-or-equal-to 3

−3<x<1

Answer 1

To solve the quadratic inequality $3x^2+3x-6\le 0$ by factoring, let's first simplify the inequality:

1. Divide the entire inequality by 3: $$x^2+x-2\le 0$$

2. We now need to factor $x^2+x-2$. We are looking for two numbers that multiply to -2 (the constant term) and add up to 1 (the coefficient of the linear term). The numbers that satisfy these conditions are 2 and -1. Thus, we can factor it as: $$(x+2)(x-1)\le 0$$

3. Next, we find the critical points by setting the factored expression equal to zero: $$(x+2)(x-1)=0$$ This gives us the critical points $x=-2$ and $x=1$.

4. Now we will test intervals around the critical points to see where the product $(x+2)(x-1)$ is less than or equal to zero. The intervals to test are:

   • $(-\mathrm{\infty },-2)$

   • $(-2,1)$

   • $(1,\mathrm{\infty })$

   • For $x<-2$ (e.g., $x=-3$): $$(-3+2)(-3-1)=(-1)(-4)=4>0\text{(not part of the solution set)}$$

   • For $-2<x<1$ (e.g., $x=0$): $$(0+2)(0-1)=(2)(-1)=-2<0\text{(part of the solution set)}$$

   • For $x>1$ (e.g., $x=2$): $$(2+2)(2-1)=(4)(1)=4>0\text{(not part of the solution set)}$$

5. Now we include the critical points because the inequality is less than or equal to zero ($\le$): $$x=-2$$ and $$x=1$$

Thus, the solution to the inequality $3x^2+3x-6\le 0$ is: $$-2\le x\le 1$$

So the correct response is: $$-2\le x\le 1$$