Easy but tough problem!

I just came across this question, which looks really easy to solve at the first sight, but took me hours to come up with nothing relevant. Here it is-

Given, if-

a2+b22ab+10=0a^2 + b^2 - 2a - b + 10 = 0, Find the value of a+ba + b.

It was a Multiple Choice Question, and hence it had options-

  • 44
  • 55
  • 66
  • 88.

Please provide with a very simple solution, as this problem doesn't seem to be the one which should be given high-level attention.

Note by Akshat Jain
6 years ago

No vote yet
2 votes

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If aa and bb are complex numbers, then we can conclude straightaway that this equation will have infinitely many solutions (four obvious solutions yielding different sums: (a,b)=(1±3i,0)(a,b)= (1 \pm 3i, 0), (a,b)=(0,1±39i2)(a,b)= \left (0, \frac{1\pm \sqrt{39}i}{2} \right ) ). If, however, aa and bb are real, we can re-arrange the equation and obtain: (a1)2+(b12)2+354=0(a-1)^2 + \left (b-\frac{1}{2} \right )^2 + \frac{35}{4} = 0 However, (a1)2+(b12)20(a-1)^2+\left (b-\frac{1}{2} \right )^2 \geq 0, which leads to the conclusion 3540 \frac{35}{4} \leq 0, an obvious contradiction.

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Shouldn't it be 354\frac{35}{4} instead of 1720\frac{17}{20}?

Lokesh Sharma - 6 years ago

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Corrected. Thanks.

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The problem seems erroneous to me. If you put a 2 behind b then the answer is 4.

Lokesh Sharma - 6 years ago

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It might be possible that there is some flaw in the question. But not considering it.

Akshat Jain - 6 years ago

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I think the options are given wrong.One can see that if we arrange the terms as:(a2a^{2} - 2a2a ++ 1010) ++ (b2b^{2} - bb) = 00. Therefore, (a2a^{2} - 2a2a ++ 1010) = 00 and (b2b^{2} - bb) = 00.Now observing the first equation we see that it's having all its roots imaginary and bb = 00 or bb = 11 . Therefore, the result is not 00 and is instead imaginary.Also, one can arrange the terms on L.H.S as (a2(a^{2} - 2a2a) ++ (b2b^{2} - bb + 1010) = 00, in this case also the same case arises an the expression is not equal to zero. One can also look as it as:(a2a^{2} - 2a2a) + (b2b^{2} - bb) = 10-10 =>=> a(2a)a(2-a) + b(1b)b(1-b) = 1010 Now, using AMGMAM \geq GM , we get 1a(2a)1 \geq a(2-a) and 14 b(1b)\frac {1}{4}\ \geq b(1-b).On adding we get, 54 10\frac {5}{4}\ \geq 10 which is a contradiction. Also,from this we find that the minimum value of the expression on the left hand side is in fact 8.75 or 354\frac {35}{4} . Hence,for no aa,bb aa + bb is equal to any of the options you have given.

Bhargav Das - 6 years ago

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My dear, (a22a+10)+(b2b)=0(a^2-2a+10) + (b^2-b)=0 doesn't imply both summands are zero. (Think of the case when one is positive & one is negative). Also don't be under the misconception that b2b>0b^2-b > 0 for all b.(Think what happens when b<1b < 1). Also AMGMAM-GM applies iff all numbers you are dealing with are positive reals.

A Brilliant Member - 6 years ago

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Thanks.I get that.

Bhargav Das - 6 years ago

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I found this question in a coaching test paper. The same exact options are given.

Akshat Jain - 6 years ago

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