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confused with this problem here

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Source: — Problem Solving |

by pemdas » Sun Mar 25, 2012 2:32 pm
cyrwr1 wrote:is this manageable?

a^2 - b^2 =30
ab=10

Can we find out what a^2+b^2 is?
IOM quadratic solution :(
a=10/b and 100/b^2 -b^2-30=0. Solve quadratics for b^2, b^4+30b^2-100=0. b^2=3.0277 and a^2=30+3.0277
a^2+b^2=36,0554
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by icanmakeit2bschool » Sun Mar 25, 2012 9:24 pm
Is it going to be in fractions ?? because we will not get the answer in decimals.
Last edited by icanmakeit2bschool on Thu Mar 29, 2012 11:21 pm, edited 1 time in total.
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by kul512 » Thu Mar 29, 2012 8:27 am
Another way to solve-

Let
X2+Y2 = A

so
X2+Y2-2XY=(X-Y)2 = A-2XY = A-20 (PUTTING VALUE OF xy=0)-----------------(1)
same way
X2+Y2+2XY=(X+Y)2 = A+2XY = A+20------------------------------------------(2)

multiplying equation 1 and 2-

(X-Y)2 * (X+Y)2 = A2-400
(X2-Y2)2 = A2-400
(30)2 = A2-400
A2=1300
so
A= 36.055
X2+Y2=36.055

Sorry i replaced a,b with X and Y...
:-)
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by ShrutiN » Thu Mar 29, 2012 11:13 am
Heres a simple solution;
Let a^2+b^2=x adding this with a^2-b^2=30 we get

2a^2=30+x
a=(30+x/2)^0.5
b=(x-30/2)^0.5

substituting these in ab=10

and squaring both sides

(x+30)(x-30)=4*100

x^2-900=400

x=a^2+b^2=(1300)^0.5 = 36.055

Hope that helps!
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by cyrwr1 » Thu Mar 29, 2012 1:41 pm
I am so confused about this too. Is that correct as the previous submitted answer of 50 seemed alright too!
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by ronnie1985 » Thu Mar 29, 2012 8:45 pm
ab = 10 => b = 10/a

a^2-b^2 = a^2 - 100/a^2 = 30 => a^4-100-30a^2 = 0 a^2 = x (let) => x = (30+/-sqrt(900+400))/2 solving we can get. a^2+b^2
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by [email protected] » Thu Mar 29, 2012 10:55 pm
this is possible but if the answer of 'k' comes out to be in decimals, this will never be asked in the GMATLAND. so do not worry, the answer has to be an integer.

Also this kind of a question can be asked in the DS question...
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by cyrwr1 » Sat Mar 31, 2012 1:19 pm
I am aware of this type of problem not on the GMAT but this is just an inquiry I had.

Is it really that confusing with a decimal response as the solution?
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by Pharo » Sat Mar 31, 2012 10:17 pm
cyrwr1 wrote:I am so confused about this too. Is that correct as the previous submitted answer of 50 seemed alright too!
Yes, square root of 1300 is the correct solution :)

Here is a different way of solving it:

a^2 - b^2 = (a-b)*(a+b) = 30

(a-b)*(a+b) = 30 ; now take the square of both sides
((a-b)*(a+b))^2 = 900
(a-b)^2 * (a+b)^2 = 900
(a^2 + b^2 - 2ab) * (a^2 + b^2 + 2ab) = 900 ; we know 2ab = 20 and let (a^2 + b^2) = x ;; sub these
(x - 20)(x + 20) = 900 ; this is like the function up there that we opened up; close this one back
x^2 - 400 = 900
x^2 = 1300 --> x = sqrt(1300) = 10sqrt(13)
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