Inequalities $- x^2 - (1/2) x - 5 < 0$ ; why is $x > 5/2$?

Question

Question :

$$\text{ find the set of values of }x \text{ for which } $$

$$10 + x - 2x^2 < 0$$

Answer :

$$x < -2$$

$$x > 2\frac{1}{2}$$

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EDIT - thanks for the responses. To try and highlight what it is I'm not understanding :

I have

$$(x+2)\left(x-\frac{5}{2}\right) > 0 $$

I now understand why this is greater than zero after factoring out the negative sign, the same way that

$$-30 < -5 $$

$$30 > 5 $$

However I'm still unsure about why we chose the answers.

As explained by mathlove -

$$\iff "x+2\gt 0\ \text{and}\ x-\frac 52\gt0"\ \text{or}\ "x+2\lt 0\ \text{and}\ x-\frac 52\lt0"$$

$$\iff x\gt \frac 52\ \ \text{or}\ \ x\lt -2.$$

I don't really see what it is that made us choose the $x < -2 $ rather than $x>-2$.... from the options above it looks like it could have been either?

Even without these options that I hadn't even considered I'm unsure about the $\frac{5}{2}$ mainly...

I'll try and demonstrate my working because I'm struggling to use words -

so from

$$(x+2)(x-\frac{5}{2}) > 0 $$

I know that

$$x-2 > 0$$

$$x+\frac{5}{2}> 0$$

So when I move the negative two over the other side of the inequality sign I change the direction

$$ x < -2 $$

and then when I effect the 5/2 I'm wrong, I feel it should go

$$x + \frac{5}{2} > 0$$

Arhh, I think a penny might have dropped -

It's because the 5/2 is positive... so

$$ x + \frac{5}{2} > 0 $$

then

$$ x < -\frac{5}{2} $$

Soo

$$ x > -\frac{5}{2}$$

is that right?

Answer 1

Note that $$10+x-2x^2\lt 0\iff \frac{10+x-2x^2}{-2}\color{red}{\gt}\frac{0}{-2}\iff x^2-\frac{1}{2}x-5\gt0$$ and that $$\left(x+2\right)\left(x-\frac 52\right)\gt0$$ $$\iff "x+2\gt 0\ \text{and}\ x-\frac 52\gt0"\ \text{or}\ "x+2\lt 0\ \text{and}\ x-\frac 52\lt0"$$ $$\iff x\gt \frac 52\ \ \text{or}\ \ x\lt -2.$$

Answer 2

You begin with $$-2x^2+x+10 < 0.$$

You divided by $-2$. But when you divide by a negative number, you must switch the inequality. Think of it like adding terms to either side:

$$-2x^2+x+10 + \color{red}{2x^2 -x - 10} < 0 + \color{red}{2x^2 -x -10}\\ 0 < 2x^2-x-10 \\ 2x^2 -x-10 > 0 $$

Answer 3

we can rewrite the inequality in the following form $10+x-2x^2<0$ is equivalent to $x^2-\frac{1}{2}x-5>0$ $(x+2)\left(x-\frac{5}{2}\right)>0$ from here we obtain $x<-2$ or $x>\frac{5}{2}$