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

Question

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.

$$(m-n)^2 = (m+n)^2 - 4mn$$ – Daniel Fischer Aug 01 '14 at 18:59 — Daniel Fischer, Aug 01 '14 at 18:59

score 10 · Answer 1 · answered Aug 01 '14 at 18:59

10

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

Can you solve this quadratic?

answered Aug 01 '14 at 18:59

JimmyK4542

mathlove · Accepted Answer · 2014-08-01T19:05:08.900

7

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

edited Aug 01 '14 at 19:05

answered Aug 01 '14 at 18:59

mathlove

What about $-1$ – Derek 朕會功夫 Aug 02 '14 at 03:10
@Derek朕會功夫: Note that what I wrote is $(m-n)^2=1$, not $m-n=1$. – mathlove Aug 02 '14 at 06:36
Oh I missed the squared part. – Derek 朕會功夫 Aug 02 '14 at 15:16

score 4 · Answer 3 · answered Aug 01 '14 at 18:59

4

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

answered Aug 01 '14 at 18:59

Nicholas Stull

score 2 · Answer 4 · edited Aug 01 '14 at 20:08

2

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

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

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

yields

$m-n=+-(1)$

edited Aug 01 '14 at 20:08

Community

answered Aug 01 '14 at 19:02

SA-255525

4 Answers4