IOM quadratic solutioncyrwr1 wrote:is this manageable?
a^2 - b^2 =30
ab=10
Can we find out what a^2+b^2 is?
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
confused with this problem here
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pemdas
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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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icanmakeit2bschool
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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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kul512
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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...
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...
Sometimes there is very fine line between right and wrong: perspective.
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!
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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ronnie1985
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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
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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[email protected]
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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...
Also this kind of a question can be asked in the DS question...
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Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
Yes, square root of 1300 is the correct solutioncyrwr1 wrote:I am so confused about this too. Is that correct as the previous submitted answer of 50 seemed alright too!
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)
