Question #5984e

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Question #5984e

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Answer 1 · 2017-11-30T07:57:01

$(x + 4) (2 x - 3) \leq 0$ $(x+4)(2x-3) <= 0$

Explanation:

We are given the quadratic inequality

$2 x^{2} + 5 x - 12 \leq 0$ $2x^2 + 5x - 12 <= 0$

Split the middle term as shown below:

$2 x^{2} + 8 x - 3 x - 12 \leq 0$ $2x^2 + 8x - 3x - 12 <= 0$

$2 x (x + 4) - 3 (x + 4) \leq 0$ $2x(x+4) - 3(x+4) <= 0$

Hence, we can conclude that

$(2 x - 3) (x + 4) \leq 0$ $(2x - 3) (x + 4) <=0$

is what we get after factorizing the given quadratic inequality.

Answer 2 · 2017-11-30T08:03:15

$x \leq \frac{3}{2}$ $x<=3/2$ and $x \leq - 4$ $x<=-4$

Explanation:

.

$2 x^{2} + 5 x - 12 \leq 0$ $2x^2+5x-12<=0$

Let's add and subtract $3 x$ $3x$ to the equation:

$2 x^{2} + 3 x - 3 x + 5 x - 12 \leq 0$ $2x^2+3x-3x+5x-12<=0$

$2 x^{2} - 3 x + 8 x - 12 \leq 0$ $2x^2-3x+8x-12<=0$

$x (2 x - 3) + 4 (2 x - 3) \leq 0$ $x(2x-3)+4(2x-3)<=0$

$(2 x - 3) (x + 4) \leq 0$ $(2x-3)(x+4)<=0$

$2 x - 3 \leq 0$ $2x-3<=0$ This gives us $x \leq \frac{3}{2}$ $x<=3/2$

$x + 4 \leq 0$ $x+4<=0$ This gives us $x \leq - 4$ $x<=-4$

Solution includes all of $(- 4, \frac{3}{2})$ $(-4,3/2)$

Answer 3 · 2017-11-30T20:53:03

Please see below.

Explanation:

$2 x^{2} + 5 x - 12 = 0$ $2x^2+5x-12 = 0$ when

$(2 x - 3) (x + 4) = 0$ $(2x-3)(x+4)=0$ which happens at

$x = \frac{3}{2}$ $x=3/2$ and at $x = - 4$ $x=-4$

These numbers partition the number line into the intervals:

$(- \infty, - 4)$ $(-oo,-4)$ $" "$ $(- 4, \frac{3}{2})$ $(-4,3/2)$ $" "$ $(\frac{3}{2}, \infty)$ $(3/2,oo)$

Pick a number in each interval and test the inequality.

I'll use $- 10$ $-10$ , $0$ $0$ and $10$ $10$

At $x = - 10$ $x = -10$ , we have
$(2 x - 3) (x + 4) = (2 (- 10) - 3) ((- 10) + 4)$ $(2x-3)(x+4)=(2(-10)-3)((-10)+4)$ which is a negative times a negative, so it is positive and the inequality is false.

At $x = 0$ $x = 0$ , we have
$(2 x - 3) (x + 4) = (2 (0) - 3) ((0) + 4)$ $(2x-3)(x+4)=(2(0)-3)((0)+4)$ which is a negative times a positive, so it is negative and the inequality is true.
The solutions include all of $(- 4, \frac{3}{2})$ $(-4,3/2)$ .

At $x = 10$ $x = 10$ , we have
$(2 x - 3) (x + 4) = (2 (10) - 3) ((10) + 4)$ $(2x-3)(x+4)=(2(10)-3)((10)+4)$ which is a positive times a positive, so it is positive and the inequality is false.

The solution set is

$(- 4, \frac{3}{2})$ $(-4,3/2)$ .

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Question #5984e

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