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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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Manager
Joined: 14 Jul 2014
Posts: 94
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.

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
Intern
Joined: 31 Jan 2016
Posts: 11
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....
Intern
Joined: 07 Jan 2017
Posts: 20
Location: India
Concentration: Other, Other
Schools: UConn"19
WE: Medicine and Health (Pharmaceuticals and Biotech)
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)
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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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?
Intern
Joined: 06 Dec 2016
Posts: 14
Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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02 Mar 2018, 08:13
Analyze: z = 2u^2 + 11
(2) u=6 => v^2 = 11- 6^2 ==> z solved
(1) y=30 => u = 30/v put it in v^2 + 11 = u^2 => (v^2)^2 + 11v^2 - 900 = 0.
Analyze: P = product of 2 roots = - 900, S = sum of roots = -11
=> Equation has 2 roots : one v^2 is positive and one v^2 is negative.
But v^2 is always positive => v^2 is only one value ==> z solved
==> D
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Joined: 04 Oct 2017
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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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02 Mar 2018, 20:22
VeritasPrepKarishma wrote:
jlgdr wrote:
Could someone please elaborate a little bit more on this problem?

How do you get to the values of (u,v) ? Is it just by ballparking? Or is there an elegant algebraic approach for this one?

Thanks
Cheers!
J

Given in the stem:

$$11 = u^2 - v^2 = (u + v)(u - v)$$
$$y = 2uv$$
$$z = u^2 + v^2$$

$$z + y = u^2 + v^2 + 2uv = (u + v)^2$$
$$z - y = u^2 + v^2 - 2uv = (u - v)^2$$

$$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$

121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11
So (z + y)(z - y) = (61 + 60)*(61 - 60) or = (11 + 0)(11 - 0)
So z can be 61 or 11 depending on whether y is 60 or 0.

Statement 1 gives y = 60. So z must be 61. Sufficient

Statement 2 gives u is 6 which means 2uv (= y) is not 0. So z must be 61. Sufficient

Answer (D)

I solved it as follows. Please let me know if my method is incorrect.

Statement 1.
y=60
uv = 30
v=30/u
substituting v=30/u in z = u2+v2 ..we get the value of u and subsequently use that value in uv = 30 to get the value of v.Hence sufficient

Statement 2
u given we know uv=30 hence sufficient.

Hence 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 May 2018, 08:23
Hi all,

I thought statement one was much easier to prove sufficient than stated above.

We know 60=2uv, 11=u^2-v^2, and Z=u^2+v^2. Thus, 30/u=v. Then plug in 30/u for v in 11=u^2-v^2---> ie 11= u^2 - (30/u)^2. There is one variable, thus we know we can solve for u. Then theoretically, we plug u in to find v. Then we would plug the solved values for u and v in Z=u^2+v^2. Thus, we could solve for z.

Can someone please correct me if this reasoning is incorrect?
Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,   [#permalink] 16 May 2018, 08:23

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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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