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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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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.

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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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]

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 ?

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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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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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)
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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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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