If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$?

If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$? I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?

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You are right and going well, and yes we can figure the values of each variable. But it is not necessary to do so :) – chubakueno Jun 3 '14 at 18:03
What happens if you expand $(a - b)^2$? – David K Jun 3 '14 at 18:05

$$\frac1{a^2}+\frac1{b^2}=4$$ $$a^2b^2\left(\frac1{a^2}+\frac1{b^2} \right)=4a^2b^2$$ $$b^2+a^2=(2ab)^2$$ $$a^2-2ab+b^2=(2ab)^2-2ab$$ $$(a-b)^2=(2ab)^2-2ab=(2\cdot3)^2-2\cdot3=30$$

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Your answer is good but try to add some words for the explanation. – Tunk-Fey Jun 3 '14 at 18:24
@Tunk-Fey - Why? each step flows well from the prior one. – JoeTaxpayer Jun 3 '14 at 19:17
@JoeTaxpayer Just in case the OP doesn't understand. You may take a look Andre Nicolas' answer. – Tunk-Fey Jun 3 '14 at 19:27
He states 'you know that $(a-b)^2=30$ ' - but it's not clear that we do. That's the question OP is asking. – JoeTaxpayer Jun 3 '14 at 19:39

Hints:

$\frac{1}{a^2}+\frac{1}{b^2} = 4 \rightarrow b^2 + a^2 = 4(ab)^2$

$(a-b)^2 = (a^2+b^2) -2(ab)$

edit:

To solve for a particular variable, you can use $ab=3 \rightarrow a = \frac{3}{b}$ to eliminate a variable. For example $a^2+b^2 = \frac{9}{b^2}+b^2$

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I already know that (a-b)² =30 so how would you figure out one of the variables – user154989 Jun 3 '14 at 18:10

You know that $(a-b)^2=30$. The same strategy tells you that $(a+b)^2=42$. Thus $$a+b=\pm \sqrt{42}\quad\text{and}\quad a-b=\pm\sqrt{30}.$$ Now by adding and subtracting, we can find $2a$ and $2b$. and hence $a$ and $b$. Note that there are $4$ combinations, though if we have found one solution $a=p$, $b=q$, the other three are $a=-p$, $b=-q$, and $a=q$, $b=p$, and $a=-q$, $b=-p$.

One of the solutions is $a=\frac{\sqrt{42}+\sqrt{30}}{2}$, $b=\frac{\sqrt{42}-\sqrt{30}}{2}$.

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This is right, of course, but it misses the point: this question is all about the symmetry in the equations, and avoiding having to solve for the values of $a$ and $b$ separately. I wish any of my high school teachers would have explained the idea that you sometimes don't care about the values of each part of an expression. – symplectomorphic Jun 3 '14 at 19:30
I was answering the question that OP had, in a comment, about solving for $a$ and $b$, given that OP already knew the value of $(a-b)^2$. The point about the solution is that it "breaks symmetry" very late in the game. Completely agree about the importance of exposure to symmetric functions. – André Nicolas Jun 3 '14 at 19:34
In fact, since I trust your judgment, Andre, do you know of a textbook that looks closely at these kinds of algebraic tricks? I think heavily symmetric equations often arise in geometry (and math competitions). But I can't think of a reference that systematically discusses problems like "if $a+b=5$ and $a^2+b^2=10$, what's the value of $ab$?" where the point isn't to find $a$ and $b$ separately. – symplectomorphic Jun 3 '14 at 19:35
I cannot think of a good source. Problem collections aimed below the Olympiad level often have symmetric examples. Algebra books used to cover this kind of material. Alas, no more. – André Nicolas Jun 3 '14 at 19:39
Alas indeed. I've written my own notes on this material and was always surprised I could never ever find a thorough, exhaustive discussion of this sort of technique. The competition books are always terse: they present a few examples but leave out the really interesting discussion of what's going on. Thanks though. – symplectomorphic Jun 3 '14 at 19:42