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$$a + b = 2\\ a^2 + b^2 = 6$$

Find the value of $(a-b)^2 $

My workings till I got stuck -

$$(a-b)^2 = a^2 - 2ab + b^2 \\ = a^2 + b^2 - 2ab\\ = 6 - 2 ab $$

I'm stuck at how to find $ab$ . Can I get hints on how to find $ab$? Thanks a lot.

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  • 3
    $\begingroup$ $4= (a+b)^2=a^2+b^2+2ab$ $\endgroup$ – lulu Feb 26 '16 at 15:16
  • $\begingroup$ Squaring the first equation then subtract the second equation from it. $\endgroup$ – Zhanxiong Feb 26 '16 at 15:17
  • $\begingroup$ The above comments mean that you should consider the quantity $2a^2+2b^2-(a+b)^2$... $\endgroup$ – abiessu Feb 26 '16 at 15:19
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Use $$(a-b)^2+(a+b)^2=2(a^2+b^2)$$ Thus, here, $$(a-b)^2+4=12$$ $$(a-b)^2=8$$

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We will solve for the values of $a$ and $b$. $a=2-b$ so $a^2+b^2=6 \iff (2-b)^2 + b^2 +6 $

This means $2b^2-4b-2=0 \iff b^2 -2b -1 +0 $. The solutions of this equation are $b=1-\sqrt{2}$ and $b=1+\sqrt{2}$. Then $a=1+\sqrt{2}$ or $a=1-\sqrt{2}$

Then, in any case $(a-b)^2=8$

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  • $\begingroup$ -1.This is a low grade method. $\endgroup$ – N.S.JOHN Apr 17 '16 at 10:42
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$$a+b=2$$

$$\implies(a+b)^2=4$$

$$\implies a^2+b^2+2ab=4$$

$$\implies2ab=-2$$

Also,

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

$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)

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