Find the value of $(a - b)^2$ given $a + b = 2$ and $a^2 + b^2 = 6$

Question

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

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.

Squaring the first equation then subtract the second equation from it. — Zhanxiong, Feb 26 '16 at 15:17
The above comments mean that you should consider the quantity $2a^2+2b^2-(a+b)^2$... — abiessu, Feb 26 '16 at 15:19

score 10 · Answer 1 · edited Feb 26 '16 at 16:06

10

Use $$(a-b)^2+(a+b)^2=2(a^2+b^2)$$ Thus, here, $$(a-b)^2+4=12$$ $$(a-b)^2=8$$

edited Feb 26 '16 at 16:06

ssinad

50

answered Feb 26 '16 at 15:23

GoodDeeds

11,185
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score 1 · Answer 2 · answered Feb 26 '16 at 15:18

1

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 $(a-b)^2=8$

answered Feb 26 '16 at 15:18

S -

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-1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42

score 1 · Answer 3 · edited Jun 12 '20 at 10:38

1

$$a+b=2$$

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

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

$$\implies2ab=-2$$

Also,

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

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

edited Jun 12 '20 at 10:38

Community

1

answered Mar 06 '16 at 12:20

Soham

9,990

Find the value of $(a - b)^2$ given $a + b = 2$ and $a^2 + b^2 = 6$

3 Answers3