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Re: GRE Math Challenge #53- If a and b are positive numbers [#permalink]
hi, can someone share the step by step solution plz?
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Re: GRE Math Challenge #53- If a and b are positive numbers [#permalink]
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Deedee wrote:
hi, can someone share the step by step solution plz?



Add both the equation::

\(a^2 + b^2 = m\)---------------I

\(a^2 -b^2 = n\) ----------II

Adding I & II

\(2a^2\) =\(m + n\)

\(a^2\)=\(\frac{(m + n)}{2}\)

Similarly substracting II from I

\(b^2\) = \(\frac{(m - n)}{2}\)

Now multiply

\(a^2 * b^2\) =\(\frac{(m^2 - n^2)}{2}\)

or \(ab\) = \(\sqrt {\frac{(m^2 -n^2)}{2}}\)
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Re: GRE Math Challenge #53- If a and b are positive numbers [#permalink]
1
I did it this way
so a^2+b^2=m so (a+b)^2=m-2ab and (a-b)^2=m+2ab
now we have a^2-b^2=n we can simplify that into (a+b)(a-b)=n
Now square the whole eq you get (a+b)^2 (a-b)^2=n^2
(m+2ab)(m-2ab)=n^2
its straight forward from here
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Re: GRE Math Challenge #53- If a and b are positive numbers [#permalink]
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Re: GRE Math Challenge #53- If a and b are positive numbers [#permalink]
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