A rad problem

Algebra Level 4

16 + 31 5 + 13 − 12 1 − ( 9 − 31 20 ) 2 \large \sqrt{\frac{16+\sqrt{31}}5 } + \sqrt{13 - 12\sqrt{1 - \left( \frac{9-\sqrt{31}}{20} \right)^2 } }

If the above expression can be represented in the form x \sqrt{x} , then find x x , where x x is square free.


The answer is 10.

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Hobart Pao by Hobart Pao , 23, USA

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2 solutions

Chew-Seong Cheong
Jul 31, 2015

16 + 31 5 + 13 − 12 1 − ( 9 − 31 20 ) 2 = 32 + 2 31 10 + 13 − 12 ( 1 + 9 − 31 20 ) ( 1 − 9 − 31 20 ) = ( 1 + 31 ) 2 10 + 13 − 12 ( 29 − 31 ) ( 11 + 31 ) 400 = 1 + 31 10 + 13 − 12 288 + 18 31 400 = 1 + 31 10 + 65 − 3 ( 3 + 3 31 ) 2 5 = 1 + 31 10 + 56 − 9 31 5 = 1 + 31 10 + 112 − 18 31 10 = 1 + 31 10 + ( 9 − 31 ) 2 10 = 1 + 31 10 + 9 − 31 10 = 10 10 = 10 \displaystyle \sqrt{\frac{16+\sqrt{31}}{5}} + \sqrt{13-12\sqrt{1-\left( \frac{9-\sqrt{31}}{20} \right)^2}} \\ \displaystyle = \sqrt{\frac{32+2\sqrt{31}}{10}} + \sqrt{13-12\sqrt{\left( 1 + \frac{9-\sqrt{31}}{20} \right) \left( 1 - \frac{9-\sqrt{31}}{20} \right)}} \\ \displaystyle = \sqrt{\frac{(1+\sqrt{31})^2}{10}} + \sqrt{13-12\sqrt{\frac{(29-\sqrt{31})(11+\sqrt{31})}{400}}} \\ \displaystyle = \frac{1+\sqrt{31}}{\sqrt{10}} + \sqrt{13-12\sqrt{\frac{288+18\sqrt{31}}{400}}} \\ \displaystyle = \frac{1+\sqrt{31}}{\sqrt{10}} + \sqrt{\frac{65- 3 \sqrt{\left(3 + 3 \sqrt{31} \right)^2}}{5}} \\ \displaystyle = \frac{1+\sqrt{31}}{\sqrt{10}} + \sqrt{\frac{56 - 9 \sqrt{31}}{5}} \\ \displaystyle = \frac{1+\sqrt{31}}{\sqrt{10}} + \sqrt{\frac{112 - 18 \sqrt{31}}{10}} \\ \displaystyle = \frac{1+\sqrt{31}}{\sqrt{10}} + \sqrt{\frac{\left(9 - \sqrt{31} \right)^2}{10}} \\ \displaystyle = \frac{1+\sqrt{31}}{\sqrt{10}} + \frac{9 - \sqrt{31}}{\sqrt{10}} = \frac{10}{\sqrt{10}} = \boxed{\sqrt{10}}

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Moderator note:

Very creative manipulation of algebraic terms!

For clarity, can you explain how you know that you should manipulate these as such?

  1. 288 + 18 31 = ( 3 + 3 31 ) 2 288 + 18\sqrt{31} = (3+3\sqrt{31})^2

  2. 56 − 9 31 5 = 112 − 18 31 10 = ( 9 − 31 ) 2 10 \large\frac{56-9\sqrt{31}}5 = \frac{112 - 18\sqrt{31}}{10} = \frac{(9-\sqrt{31})^2}{10}

  3. 16 + 31 5 = 32 + 2 31 10 = ( 1 + 31 ) 2 10 \large\frac{16+\sqrt{31}}5 = \frac{32 + 2\sqrt{31}}{10} = \frac{(1+\sqrt{31})^2}{10}

Lu Chee Ket
Oct 17, 2015

(b^2 - 4 a c) as square of whole number as the proper way. However, by using calculator, value for Sqrt (10) is found by a square. Therefore x = 10 is found by technology without hard.

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