This cannot happen

Today I was going to post a problem but i suddenly got confused about problem.The problem was if 14132+23528+3=a\sqrt { \sqrt { 14-\sqrt { 132 } } +\sqrt { 23-\sqrt { 528 } + } \sqrt { 3 } } =a.Find the value of aa.

My solution is following:

14132=3+112132=(311)2 \sqrt { 14-\sqrt { 132 }}=3+11-2\sqrt { 132 }=\sqrt { { \left( \sqrt { 3 } -\sqrt { 11 } \right) }^{ 2 } }

Similarly

23528=(1112)2\sqrt { 23-\sqrt { 528 } }=\sqrt { { \left( \sqrt { 11 } -\sqrt { 12 } \right) }^{ 2 } }

And 3=1512=12+36×2=122+32236=(123)2\sqrt { 3 }=\sqrt {15-12}=\sqrt { 12+3-6\times 2}=\sqrt { \sqrt { { 12 }^{ 2 } } +\sqrt { { 3 }^{ 2 } } -2\sqrt { 36 }} =\sqrt { { \left( \sqrt { 12 } -\sqrt { 3 } \right) }^{ 2 } }

Therefore

14132+23528+3=(311)2+(1112)2+(123)2=0\sqrt { \sqrt { 14-\sqrt { 132 } } +\sqrt { 23-\sqrt { 528 } + } \sqrt { 3 } } =\sqrt { { \left( \sqrt { 3 } -\sqrt { 11 } \right) }^{ 2 } } +\sqrt { { \left( \sqrt { 11 } -\sqrt { 12 } \right) }^{ 2 } } +\sqrt { { \left( \sqrt { 12 } -\sqrt { 3 } \right) }^{ 2 } } =0

But now use calculator and evaluate i get answer above 0.

Please help me about this.Which step did I do wrong.

Note by Shivamani Patil
5 years ago

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1 vote

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Convention is that x\sqrt { x } returns a positive value, even though it can be both positive and negative. Hence, if we rewrite your expression in this way

(113)2+(1211)2+(123)2\sqrt { { (\sqrt { 11 } -\sqrt { 3 } ) }^{ 2 } } +\sqrt { { (\sqrt { 12 } -\sqrt { 11 } ) }^{ 2 } } +\sqrt { { (\sqrt { 12 } -\sqrt { 3 } ) }^{ 2 } }

you'll get a non-zero value which agrees with the original expression.

This is one of the big problems with the idea that a "function can only return one value". By insisting that functions can "only return one value", we're missing out the big picture. It's an artificial restriction. I prefer using implicit equations for exactly that reason, it takes care of all the problems with signs. For example, y=1x2y=\sqrt { 1-{ x }^{ 2 } } is not even a circle, it's only a half of a circle. But y2=1x2{ y }^{ 2 }=1-{ x }^{ 2 } is a circle.

Michael Mendrin - 5 years ago

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How did you get above expression?

shivamani patil - 5 years ago

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It's exactly the same as yours, except that I reversed the order of some of the terms, which I can do since, after being squared, it's the same. That is, for example, in the 2nd line, you could have reversed the terms of the expression on the far right with equal mathematical validity.

Michael Mendrin - 5 years ago

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@Michael Mendrin If i want to post this problem then what conditions i should mention to make it valid problem.

shivamani patil - 4 years, 11 months ago

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@Shivamani Patil A calculator will always return a postive square root. Roank above has given a good explanation.

Michael Mendrin - 4 years, 11 months ago

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The big mistake was x2=x\sqrt{{x}^{2}}=\left| x \right| and not xx. Hope this clarifies things a bit.

Ronak Agarwal - 5 years ago

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