# Evaluating absolute inequalities

I have the *expression $$\frac{1}{\sqrt{4a^2-b^2}}$$

and I am being asked to evaluate the case when $2|a| \geq|b|$

Logically I know what this statement mean but don't know how it applies to this problem: Cases when $a$ is twice greater than the distance of $b$ is from $0$. (Correct me if i'm wrong)

I also know that absolute values on both sides of an equation are inpracticle and it would be better to write the inequality as $|a/b| \geq 1/2$.

Which can also be written as $a/b \geq 1/2$ or $a/b \leq 1/2$ which doesn't make any sense.

here's a link to the problem (#3) http://imgur.com/a/GheWI

Thank you, I would appreciate help I've been struggling with this for a while

• That first line is not an equation. What equals what? – lulu Jan 12 '17 at 16:16
• My problem dosent set it up as an equation – Olivier Perrault Jan 12 '17 at 16:17
• In that case, it's an expression. – TastyRomeo Jan 12 '17 at 16:20
• Then why do you call it an equation? Do you just mean that you want to understand the function $F(a,b)=\frac 1{\sqrt {4a^2-b^2}}$ ? what do you want to know about that function? – lulu Jan 12 '17 at 16:20
• I guess it is not a function but an expression – Olivier Perrault Jan 12 '17 at 16:23

we have $$\frac{\sqrt{4a^2-b^2}}{-(4a^2-b^2)}=-\frac{1}{\sqrt{4a^2-b^2}}$$ and here we have $$2|a|>|b|$$
The only reason they say $2|a| \geq|b|$ is that if $2|a| \not\geq|b|$, the square root $\sqrt{4a^2 - b^2}$ doesn't make sense.
• Sure. $$4a^2-b^2\geq0\\4a^2\geq b^2\\2|a|\geq|b|$$ – Arthur Jan 12 '17 at 18:49