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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
4 years 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.

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

- 4 years ago

Corrected. Thanks.

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

- 4 years ago

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

- 4 years ago

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.

- 4 years ago

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.

- 4 years ago

Thanks.I get that.

- 4 years ago

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

- 4 years ago