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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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30 Mar 2015, 08:54

It seems that the only exact method to find u and v is solving for quadratic (4th power) equation, but it's also too time consuming. If someone knows quick method of finding u and v without assuming that tose numbers are integers please share it.[/quote]

If \(x = u^2 - v^2\), \(y = 2uv\) and \(z = u^2 + v^2\), and if \(x = 11\), what is the value of z?

my approach towards statement 2: x=11 and u=6, thus 11=6^2-v^2 25=v^2 v=5 or v=-5 Since z=u^2+v^2, it does not matter if v is 5 or -5, so z=55 Please correct me if something is wrong!/spoiler] I do not understand statement 1, please help me

Responding to a pm:

Quote:

I have a quick question about how ( u + v)^2 * ( u - v)^2 = 11^2 in the problem below. If (u - v)^2 = (u + v) * (u - v) = 11 , then does ( u + v)^2 * ( u - v)^2 = 11^2 ? Is this correct? If not, can you please help me understand this?

Note that \(x = u^2 - v^2\) and x is given as 11.

So, u^2 - v^2 = 11 \((u + v)*(u - v) = 11\)

So \([(u + v)*(u - v)]^2 = 11^2\) \((u+v)^2 * (u - v)^2 = 11^2\)

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