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Finding the value of $m-n$ if $m+n=7$, $mn=12$

I tried the following,

As we know, $a^2+b^2=(a-b)^2+2ab$

But, where do I put $m+n$ value?

Please help. I am stuck.

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8  
$$(m-n)^2 = (m+n)^2 - 4mn$$ –  Daniel Fischer Aug 1 at 18:59

4 Answers 4

up vote 2 down vote accepted

You can use $$(m-n)^2=(m+n)^2-4mn=7^2-4\cdot 12=1.$$ So?

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What about $-1$ –  Derek 朕會功夫 Aug 2 at 3:10
    
@Derek朕會功夫: Note that what I wrote is $(m-n)^2=1$, not $m-n=1$. –  mathlove Aug 2 at 6:36
    
Oh I missed the squared part. –  Derek 朕會功夫 Aug 2 at 15:16

Hint: Since $m+n = 7$, we have $m = 7-n$. Hence, $12 = mn = (7-n)n$.

Can you solve this quadratic?

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Go one step further with what you are trying to use: $$(a+b)^2 = (a-b)^2 + 4ab$$

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$(m+n)^2-(m-n)^2=4mn$

$7^2-(m-n)^2=4*12$

$49-48=(m-n)^2$

yields

$m-n=+-(1)$

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1  
mn = 12, not 14 –  Nicholas Stull Aug 1 at 19:02

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