How to solve an inequality with a fraction and a negative number?

[basty]

How do you solve x for the below inequality?

$\frac{a}{x^2} < -b$

My attempt is:

$\frac{a}{x^2} + b < 0$

$\frac{a + bx^2}{x^2} < 0$

 

[andrewkirk]

No. If it asks you to solve for x, it means you need to get x by itself on one side of the (in)equation. How would you go about doing that?

 

[basty]

No. If it asks you to solve for x, it means you need to get x by itself on one side of the (in)equation. How would you go about doing that?

Do you mean it has no solution?

Can I do such as below?

$\frac{a}{x^2} < -b$

$a < -bx^2$

$-\frac{a}{b} > x^2$

$x < ±\sqrt{-\frac{a}{b}}$

$x < ±i\sqrt{\frac{a}{b}}$

$x < ±i\frac{\sqrt{ab}}{b}$

 

[davidmoore63@y]

You have to be more careful than that. Split in into 4 cases according to whether a and b are positive or negative. And a fifth case if b=0.

 

[Mark44]

You have to be more careful than that. Split in into 4 cases according to whether a and b are positive or negative. And a fifth case if b=0.

The four cases can be reduced to two: ab > 0 and ab < 0.

 

[HallsofIvy]

The four cases can be reduced to two: ab > 0 and ab < 0.

No, they can't. For example if a and b are both equal to -1, the inequality is $-\frac{1}{x^2}< 1$. Multiplying both sides by the positive $x^2$, $-1< x^2$ which is true for all x. But if a and b are both equal to 1, we have $\frac{1}{x^2}< -1$ so that $1< -x^2$ and multiplying both sides by negative 1, $-1> x^2$ which is never true.