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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-

\(a^2 + b^2 - 2a - b + 10 = 0\), Find the value of \(a + b\).

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

  • \(4\)
  • \(5\)
  • \(6\)
  • \(8\).

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
3 years, 5 months ago

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If \(a\) and \(b\) are complex numbers, then we can conclude straightaway that this equation will have infinitely many solutions (four obvious solutions yielding different sums: \((a,b)= (1 \pm 3i, 0)\), \((a,b)= \left (0, \frac{1\pm \sqrt{39}i}{2} \right ) \)). If, however, \(a\) and \(b\) are real, we can re-arrange the equation and obtain: \[(a-1)^2 + \left (b-\frac{1}{2} \right )^2 + \frac{35}{4} = 0\] However, \((a-1)^2+\left (b-\frac{1}{2} \right )^2 \geq 0\), which leads to the conclusion \( \frac{35}{4} \leq 0\), an obvious contradiction. Sreejato Bhattacharya · 3 years, 5 months ago

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@Sreejato Bhattacharya Shouldn't it be \(\frac{35}{4}\) instead of \(\frac{17}{20}\)? Lokesh Sharma · 3 years, 5 months ago

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@Lokesh Sharma Corrected. Thanks. Sreejato Bhattacharya · 3 years, 5 months ago

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The problem seems erroneous to me. If you put a 2 behind b then the answer is 4. Lokesh Sharma · 3 years, 5 months ago

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@Lokesh Sharma It might be possible that there is some flaw in the question. But not considering it. Akshat Jain · 3 years, 5 months ago

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I think the options are given wrong.One can see that if we arrange the terms as:(\(a^{2}\) \(-\) \(2a\) \(+\) \(10\)) \(+\) (\(b^{2}\) \(-\) \(b\)) = \(0\). Therefore, (\(a^{2}\) \(-\) \(2a\) \(+\) \(10\)) = \(0\) and (\(b^{2}\) \(-\) \(b\)) = \(0\).Now observing the first equation we see that it's having all its roots imaginary and \(b\) = \(0\) or \(b\) = \(1\) . Therefore, the result is not \(0\) and is instead imaginary.Also, one can arrange the terms on L.H.S as \((a^{2}\) \(-\) \(2a\)) \(+\) (\(b^{2}\) \(-\) \(b\) + \(10\)) = \(0\), in this case also the same case arises an the expression is not equal to zero. One can also look as it as:(\(a^{2}\) \(-\) \(2a\)) + (\(b^{2}\) \(-\) \(b\)) = \(-10\) \(=>\) \(a(2-a)\) + \(b(1-b)\) = \(10\) Now, using \(AM \geq GM\) , we get \(1 \geq a(2-a)\) and \(\frac {1}{4}\ \geq b(1-b)\).On adding we get, \(\frac {5}{4}\ \geq 10\) 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 \(\frac {35}{4}\) . Hence,for no \(a\),\(b\) \(a\) + \(b\) is equal to any of the options you have given. Bhargav Das · 3 years, 5 months ago

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@Bhargav Das My dear, \((a^2-2a+10) + (b^2-b)=0\) 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 \(b^2-b > 0\) for all b.(Think what happens when \(b < 1\)). Also \(AM-GM \) applies iff all numbers you are dealing with are positive reals. Paramjit Singh · 3 years, 5 months ago

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@Paramjit Singh Thanks.I get that. Bhargav Das · 3 years, 5 months ago

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@Bhargav Das I found this question in a coaching test paper. The same exact options are given. Akshat Jain · 3 years, 5 months ago

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