Polynomial of Root

Algebra Level 3

28 + 10 3 + 28 − 10 3 + 6 + 4 2 + 6 − 4 2 = ? \sqrt{28+10\sqrt3} + \sqrt{28 - 10\sqrt3} + \sqrt{6 + 4\sqrt2} + \sqrt{6-4\sqrt 2} = \ ?


The answer is 14.

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Dhira Tengara by Dhira Tengara , 19, Indonesia

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3 solutions

Let 28 + 10 3 + 28 − 10 3 = x \sqrt{28+10\sqrt{3}} + \sqrt{28-10\sqrt{3}}\ = x and 6 + 4 2 + 6 − 4 2 = y \sqrt{6+4\sqrt{2}} + \sqrt{6-4\sqrt{2}}\ = y

x 2 = 28 + 10 3 + 2 2 8 2 − 1 0 2 ( 3 ) + 28 − 10 3 x 2 = 56 + 2 484 x 2 = 56 + 2 ( 22 ) x 2 = 100 x = 10 x^2 = 28+10\sqrt{3} + 2\sqrt{28^2-10^2(3)} + 28 - 10\sqrt{3} \\x^2 = 56 + 2\sqrt{484} \\x^2 = 56 +2(22) \\x^2 = 100 \\\boxed{x= 10}

y 2 = 6 + 4 2 + 2 6 2 − 4 2 ( 2 ) + 6 − 4 2 y 2 = 12 + 2 4 y 2 = 12 + 2 ( 2 ) y 2 = 16 y = 4 y^2 = 6+4\sqrt{2} + 2\sqrt{6^2-4^2(2)} + 6 - 4\sqrt{2} \\y^2 = 12 + 2\sqrt{4} \\ y^2 = 12 +2(2) \\y^2 = 16 \\\boxed{y = 4}

x + y = 10 + 4 = 14 x+y = 10+4 = \large{\boxed{14}}

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Moderator note:

Like Dhira Tengara's solution, you didn't show that all the expressions inside their respective radicals are non-negative. It is important because the answer would turn out wrong if we were to solve − 1 + − 4 + − 9 + − 16 \sqrt{-1} + \sqrt{-4} + \sqrt{-9} + \sqrt{-16} .

It's obvious that 28 + 10 3 > 0 28 + 10\sqrt3 > 0 because it is the sum of two positive terms, but it is not obvious that 28 − 10 3 > 0 28 - 10\sqrt3 > 0 . Consider the square of 28 28 and 10 3 10\sqrt3 which are 784 784 and 300 300 respectively. Thus ( 28 + 10 3 ) ( 28 − 10 3 ) = 784 − 300 > 0 ( \sqrt{28} + 10\sqrt3)(\sqrt{28} - 10\sqrt3) = 784 - 300 > 0 then either 28 + 10 3 , 28 − 10 3 28 + 10\sqrt3, 28 - 10\sqrt3 are both negative or both positive. Since we have shown that one of them is positive, the other must be positive as well.

Can you prove that 6 − 4 2 > 0 6 - 4\sqrt2 > 0 from here?

Sorry about that.

To prove that 6 − 4 2 6-4\sqrt{2} is positive we take the square of 6 6 and square of 4 2 4\sqrt{2} which give 36 36 and 32 32 respectively. By performing difference of two squares we get that 36 − 32 = 4 36-32 = 4 , which is a positive number. From a 2 − b 2 = ( a + b ) ( a − b ) a^2 - b^2 = (a+b)(a-b) its factors are ( 6 + 4 2 ) (6 + 4\sqrt{2}) and ( 6 − 4 2 ) (6 - 4\sqrt{2}) and to get a positive product, you need the two factors to be both positive or both negative. Since 6 + 4 2 6 + 4\sqrt{2} is positive, 6 − 4 2 6 - 4\sqrt{2} is positive also.

Justine Allen Arciga - 6 years ago

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Another way to prove that it's nonnegative.

10 3 = 300 10 \sqrt{3} = \sqrt{300}

324 > 300 \sqrt {324} > \sqrt{300}

18 > 300 18 > \sqrt {300}

28 > 18 > 300 28 > 18 > \sqrt {300}

So 28 − 10 3 28 - 10 \sqrt{3} is positive.

Likewise, 4 2 = 32 4 \sqrt{2} = \sqrt{32}

36 > 32 \sqrt {36} > \sqrt {32}

6 > 32 6 > \sqrt {32}

So 6 − 4 2 6 - 4\sqrt{2} is positive also.

Justine Allen Arciga - 6 years ago
Dhira Tengara
Jun 1, 2015

All the expressions inside the radicals are all non-negative.

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Moderator note:

Almost correct. You need to make sure that all the expressions inside the radicals are all non-negative, else you would be dealing with complex numbers. And it would be better to clarify how you move from step 1 to step 2 of your solution.

Ins response to Dhira Tengara: Just make sure you have written out all the details :

28 + 10 3 = 25 2 + 3 2 + 2 ⋅ 25 ⋅ 3 \sqrt{28 + 10\sqrt{3}} = \sqrt{\sqrt{25}^{2} + \sqrt{3}^{2} + 2\cdot \sqrt{25}\cdot \sqrt{3}}

In response to the Challenge Master:

Wow! There are a lot of Challenge Master notes these days . Superb!

A Former Brilliant Member - 6 years ago
Vadlamani Madhav
Jun 16, 2015

By taking square root of that polynomial we get 5+5+2+2=14 and other terms get cancelled.

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