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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, 07:54
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, 22:49
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

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, 11:44
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, 17:57
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, 17:52
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]

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New post 02 Mar 2018, 07:13
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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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 02 Mar 2018, 19:22
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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New post 16 May 2018, 07:23
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?
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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 06 Nov 2018, 19:07
Bunuel 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

@sethiprashant i also think you are right with your reasoning that u and v do not have to be integers and thus 6 and 5 are not the only possible combination. However, the OA is D... :/
What are we overlooking?

Just to prove that i quote the question correctly, i attached a screenshot of the question


For (1) we have:
\(11 = u^2 - v^2\);

\(60= 2uv\);

and we need to find the value of \(u^2 + v^2\).

If you solve \(11 = u^2 - v^2\) and \(60= 2uv\), you get (u, v) = (-6, -5) or (u, v) = (6, 5). In either case \(u^2 + v^2=36+25\).

Thus the answer is D.

Hope it's clear.


Can sb help me understand why (u, v) = (-6, -5) or (u, v) = (6, 5) can't be also (-30,-1) or (30,1)?
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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 09 Dec 2018, 11:58
What information do we know?
There are 5 variables \(x,y,z,u,v\). The question gives 4 equations so we should only need one more equation to solve for \(z\).

Cond 1) \(y = 60, y=2uv \implies 30 = uv \implies z = u^2 + v^2 = u^2 + (\frac{30}{u})^2 \implies u^4 - 11u^2 - 900 = 0.\)
We know that this equation has the form \((u^2 - Q)(u^2 +R) = 0\) for two numbers \(Q,R\). (To verify this is actually the case we can verify that the Determinant > 0 (i.e. \(D = b^2 - 4ac > 0, \text{ where quadratic is given by } ax^2 + bx + c\)). Next, since \(u^2\) is always positive, we know that the answer has to be \(u^2 = Q\).

Since we know \(x, u^2\), we know \(v^2\). Since we now know \(v^2\) and \(u^2\) we now know \(z\).

Sufficient.

Cond 2) Since we know \(u\), we know \(u^2\). Since we know \(x, u^2\), we know \(v^2\). Since we now know \(v^2\) and \(u^2\) we now know \(z\).

Sufficient.

(D)
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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 10 Dec 2018, 01:29
Hi Bunuel, VeritasKarishma and chetan2u

I took a different approach to prove that S(1) is sufficient...
I would appreciate it if you could confirm my logic:)

We know \(11=(U+V)*(U-V)\)

And we are given that \(30=U*V\), so \(U=\frac{30}{V}\)

Plugging that into the equation above we get \(11=\frac{30^2}{V^2}-V^2\), which simplifies into \(V^4+11^2-900=0\) then I replaced \(V^2\) in the equation with \(A\), so \(V^2=A\)

\(A^2+11A-900=0\) --> \((A+36)*(A-25)\) so \(V^2\) can bei either \(-36\) or \(25\), but since a power of \(2\) can never result in a negative number \(V^2=25\)

Is my reasoning above correct? Thank you in advance!
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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 Dec 2018, 01:52
2
T1101 wrote:
Hi Bunuel, VeritasKarishma and chetan2u

I took a different approach to prove that S(1) is sufficient...
I would appreciate it if you could confirm my logic:)

We know \(11=(U+V)*(U-V)\)

And we are given that \(30=U*V\), so \(U=\frac{30}{V}\)

Plugging that into the equation above we get \(11=\frac{30^2}{V^2}-V^2\), which simplifies into \(V^4+11^2-900=0\) then I replaced \(V^2\) in the equation with \(A\), so \(V^2=A\)

\(A^2+11A-900=0\) --> \((A+36)*(A-25)\) so \(V^2\) can bei either \(-36\) or \(25\), but since a power of \(2\) can never result in a negative number \(V^2=25\)

Is my reasoning above correct? Thank you in advance!


Yes, you are correct..
v would give you two values and accordingly u will take two values.
However z will come out same in both cases.

Of course you don't have to find each case...
z=u^2+v^2...u^2=11+v^2, so z=11+v^2+v^2=11+2v^2

Your reasoning is 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]

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New post 04 Jan 2019, 04:07
without solving for the values i chose to count the variables and the equations and surprisingly my ans is right.Please correct me if i am wrong
x = u^2 – v^2,
y = 2uv,
z = u^2 + v^2
X=11

1. y=60
11= u^2 – v^2, since x=11
60= 2uv, since y = 60
z = u^2 + v^2
no. of variables (u, v and z) is equal to number of equation.Hence Sufficient.

2.U=6
Same as above.Hence answer is D
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, &nbs [#permalink] 04 Jan 2019, 04:07

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