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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

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    $\begingroup$ That first line is not an equation. What equals what? $\endgroup$ – lulu Jan 12 '17 at 16:16
  • $\begingroup$ My problem dosent set it up as an equation $\endgroup$ – Olivier Perrault Jan 12 '17 at 16:17
  • $\begingroup$ In that case, it's an expression. $\endgroup$ – TastyRomeo Jan 12 '17 at 16:20
  • $\begingroup$ 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? $\endgroup$ – lulu Jan 12 '17 at 16:20
  • $\begingroup$ I guess it is not a function but an expression $\endgroup$ – Olivier Perrault Jan 12 '17 at 16:23
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we have $$\frac{\sqrt{4a^2-b^2}}{-(4a^2-b^2)}=-\frac{1}{\sqrt{4a^2-b^2}}$$ and here we have $$2|a|>|b|$$

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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.

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  • $\begingroup$ because it would make it negative? $\endgroup$ – Olivier Perrault Jan 12 '17 at 16:24
  • $\begingroup$ @OlivierPerrault Yes, it would make the expression inside the square root negative. $\endgroup$ – Arthur Jan 12 '17 at 16:26
  • $\begingroup$ Is there a full proof way to come up with conditions to make sure that the root is not negative like this? $\endgroup$ – Olivier Perrault Jan 12 '17 at 17:50
  • $\begingroup$ Sure. $$4a^2-b^2\geq0\\4a^2\geq b^2\\2|a|\geq|b|$$ $\endgroup$ – Arthur Jan 12 '17 at 18:49

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