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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

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]

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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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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

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?
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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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06 Nov 2018, 20:07
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 $$u^2 + v^2=36+25$$.

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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09 Dec 2018, 12:58
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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10 Dec 2018, 02:29

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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10 Dec 2018, 02:52
2
T1101 wrote:

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

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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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04 Jan 2019, 05:07
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,  [#permalink]

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24 Feb 2019, 10:45
1
Top Contributor
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

Target question: What is the value of z?

Given:
u² - v² = 11
y = 2uv
z = u² + v²

Statement 1: y = 60
We can write: 60 = 2uv
Divide both sides by 2 to get: 30 = uv
Solve for u to get: u = 30/v

Take: u² - v² = 11
Replace u with 30/v to get: (30/v)² - v² = 11
Simplify to get: 900/v² - v² = 11
Multiply both sides by v² to get: 900 - v⁴ = 11v²
Rearrange to get: v⁴ + 11v² - 900 = 0
Factor to get: (v² + 36)(v² - 25) = 0
So, EITHER v² + 36 = 0 OR v² - 25 = 0
If v² + 36 = 0, then there is NO SOLUTION
If v² - 25 = 0, then v = 5 or v = -5

Statement 1 indirectly tells us that u = 30/v
If v = 5, then u = 6, in which case the answer to the target question is z = u² + v² = 6² + 5² = 61
If v = -5, then u = -6, in which case the answer to the target question is z = u² + v² = (-6)² + (-5)² = 61
NOTE, for both possible cases, we get the SAME answer to the target question
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: u = 6
We know that u² - v² = 11
Replace u with 6 to get: 6² - v² = 11
Simplify to get: 36 - v² = 11
Keep going: -v² = -25
So: v² = 25
So, EITHER v = 5 OR v = -5

Let's check each possible case:
Case a: u = 6 and v = 5. In this case, the answer to the target question is z = u² + v² = 6² + 5² = 61
Case b: u = 6 and v = -5. In this case, the answer to the target question is z = u² + v² = 6² + (-5)² = 61
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Cheers,
Brent
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,   [#permalink] 24 Feb 2019, 10:45

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