Question

Розв'яжіть нерівність (Решить неравенство) - 2x^2>=x-6

Answer 1

$-2x^{2} \geqx-6\\\\-2x^{2}-x+6\geq0\\\\2x^{2}+x-6\leq 0\\\\2x^{2}+x-6=0\\\\D=1^{2}-4*2*(-6)=1+48=49=7^{2}\\\\x_{1} =\frac{-1-6}{4}=-\frac{7}{4}=-1,75\\\\x_{2}=\frac{-1+6}{4}=\frac{5}{4}=1,25\\\\2x^{2}-x+6=2(x+1,75)(x-1,25)\\\\2(x+1,75)(x-1,25) \leq0\\\\(x+1,75)(x-1,25) \leq 0$

    +                      -                      +

______[- 1,75]______[1,25]_______

              //////////////////////////

Ответ : x ∈ {- 1,75 ; 1,25]

Answer 2

Sure, let's solve this inequality step by step.

The given inequality is:

$-2x^2 \geq x - 6$

First, let's bring all the terms to one side to set the inequality to zero:

$-2x^2 - x + 6 \geq 0$

Now, let's factor the quadratic expression:

$-(2x^2 + x - 6) \geq 0$

Factoring gives:

$-(2x - 3)(x + 2) \geq 0$

Now, we need to find the critical points by setting each factor equal to zero:

$2x - 3 = 0 \Rightarrow x = \frac{3}{2}$

$x + 2 = 0 \Rightarrow x = -2$

These critical points divide the number line into three intervals: $(- \infty, -2)$, $(-2, \frac{3}{2})$, and $(\frac{3}{2}, \infty)$.

Now, we can test each interval by choosing a test point in each interval and checking the sign of the expression $-(2x - 3)(x + 2)$.

1. Test $x = -3$ in $(- \infty, -2)$:

   $-(2(-3) - 3)(-3 + 2) = -15$

   Since it's negative, this interval is part of the solution.

2. Test $x = 0$ in $(-2, \frac{3}{2})$:

   $-(2(0) - 3)(0 + 2) = 6$

   Since it's positive, this interval is not part of the solution.

3. Test $x = 2$ in $(\frac{3}{2}, \infty)$:

   $-(2(2) - 3)(2 + 2) = -10$

   Since it's negative, this interval is part of the solution.

So, the solution to the inequality is:

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