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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,

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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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New post 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?

Given: \(u^2 - v^2=11\) and \(y = 2uv\).
Question: \(u^2 + v^2=?\)

(1) y = 60 --> \(2uv=60\) --> \(4u^2v^2=3,600\).

Square \(u^2 - v^2=11\): \(u^4-2u^2v^2+v^2=121\);

Add \(4u^2v^2\) to both sides: \(u^4+2u^2v^2+v^4=121+3,600\);

Apply \(a^2+2ab+b^2=(a+b)^2\): \((u^2+v^2)^2=121+3,600\).

\(u^2+v^2=\sqrt{121+3,600}=61\).

Sufficient.

(2) u = 6 --> \(36 - v^2=11\) --> \(v^2=25\) --> \(u^2+v^2=36+25=61\). Sufficient.

Answer: D.

Hope it helps.[/quote]


Hi Bunuel

Stmtn 1
Can we use Differences of Square approach

Property - The difference between squares grows as the squares themselves get larger

Hence given u^2 - v^2 = x = 11

Thus the only numbers that fit this equation are when u = 6 and v = 5. Hence we can find Z - Sufficient

Similalry Stmtn 2 = Suff

Hence D

Pls mention if this approach is correct

Thanks

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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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New post 10 Feb 2016, 23:49
vogelleblanc wrote:
If \(x = u^2 - v^2\), \(y = 2uv\) and \(z = u^2 + v^2\), and if \(x = 11\), what is the value of z?

(1) y = 60
(2) u = 6

[Reveal] Spoiler:
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 \([(u + v)*(u - v)]^2 = 11^2\)
\((u+v)^2 * (u - v)^2 = 11^2\)

Does this help?
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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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New post 14 Mar 2016, 12:44
kinjiGC wrote:
mau5 wrote:
vogelleblanc wrote:
If \(x = u^2 - v^2\), \(y = 2uv\) and \(z = u^2 + v^2\), and if \(x = 11\), what is the value of z?

(1) y = 60
(2) u = 6



Another approach, for F.S 1, in which you don't have to calculate for anything

We know that \(x = u^2 - v^2\) = 11,Squaring on both sides, we have \(u^4 + v^4 - 2*v^2*u^2\) = 121

Thus, adding\(4*v^2*u^2\) on both sides, we have \((u^2 + v^2)^2 = 121+4*v^2*u^2 = z^2\)

Hence, this statement is sufficient to calculate the value of z.


As Z can be negative as well as +ve, Z value cannot be uniquely determined by using the above method.

It is better to solve equations and determine.


Z can't be negative as its a sum of squares....

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

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New post 17 Feb 2017, 18:57
x=u^2−v^2..........(1).......x=(u+v)(u-v).........(u+v)=x/(u-v)
y=2uv.........(2)
z=u^2+v^2.......(3)
Adding eqn (2)+(3) results.......
z+y=u^2+v^2+2uv=>>>>>(u+v)^2=>>>>((x/(u-v))^2=>>>>>x^2/u^2-2uz+v^2
z+y=x^2/z-y
(z+y)(z-y)=x^2
z^2-y^2=x^2

(1) y = 60
z^2-y^2=x^2
z^2-60^2=11^2
z^2=60^2+11^2
we get the value of z

(2) u = 6
From eqn (1) we get the value of v and substituting v in eqn (2) we get y
using z^2-y^2=x^2 we get the value of z.

Ans.D

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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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New post 16 Aug 2017, 18:52
If u^2 - v^2 = 11
--> difference of squares
--> 36 - 25 = 11
--> z = u^2 + v^2 = 36 + 25

Thus z is known before either statements 1 or 2. Correct?

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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,   [#permalink] 16 Aug 2017, 18:52

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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,

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