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Algebra Level 3

If 1 2 + 0 + 1 + 5 \frac{1}{\sqrt{2} + \sqrt{0} + \sqrt{1} + \sqrt{5}} can be written as x − y ( z + 1 ) 2 + 2 \frac{x - \sqrt{y}(z + 1)}{2} + \sqrt{2}

Then find x − y + z 2 x - y + {z}^{2}


The answer is 0.

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Kartik Sharma by Kartik Sharma , 21, India

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3 solutions

1 2 + 0 + 1 + 5 = 1 5 + 2 + 1 = 5 + 2 − 1 ( 5 + 2 ) 2 − 1 \dfrac {1}{\sqrt{2}+\sqrt{0} + \sqrt{1}+\sqrt{5} } = \dfrac {1}{\sqrt{5}+\sqrt{2} + 1} = \dfrac {\sqrt{5}+\sqrt{2} - 1}{(\sqrt{5}+\sqrt{2})^2 - 1}

= 5 + 2 − 1 7 + 2 10 − 1 = 5 + 2 − 1 2 10 + 6 = ( 5 + 2 − 1 ) ( 2 10 − 6 ) 40 − 36 = \dfrac {\sqrt{5}+\sqrt{2} - 1}{7+2\sqrt{10} - 1} = \dfrac {\sqrt{5}+\sqrt{2} - 1}{2\sqrt{10}+6} = \dfrac {(\sqrt{5}+\sqrt{2} - 1)(2\sqrt{10}-6)}{40-36}

= 2 50 + 2 20 − 2 10 − 6 5 − 6 2 + 6 4 = \dfrac {2\sqrt{50}+2\sqrt{20} - 2\sqrt{10}-6\sqrt{5}-6\sqrt{2} + 6}{4}

= 50 + 20 − 10 − 3 5 − 3 2 + 3 2 = \dfrac {\sqrt{50}+\sqrt{20} - \sqrt{10}-3\sqrt{5}-3\sqrt{2} + 3}{2}

= 5 2 + 2 5 − 10 − 3 5 − 3 2 + 3 2 = \dfrac {5\sqrt{2}+2\sqrt{5} - \sqrt{10}-3\sqrt{5}-3\sqrt{2}+3}{2}

= 3 + 2 2 − 5 − 10 2 = 3 − 5 ( 2 + 1 ) 2 − 2 = \dfrac {3+2\sqrt{2}-\sqrt{5} - \sqrt{10}}{2} = \dfrac {3-\sqrt{5}(\sqrt{2}+1)}{2} - \sqrt{2}

⇒ x = 3 \Rightarrow x=3 , y = 5 y=5 and z = 2 ⇒ x − y + z 2 = 0 z=\sqrt{2}\quad \Rightarrow x-y+z^2 = \boxed{0}

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Aareyan Manzoor
Jan 5, 2015

A = 1 2 + 5 + 1 = 2 − 5 + 1 ( 2 + 5 + 1 ) ( 2 − 5 + 1 ) = 2 − 5 + 1 − 2 + 2 2 A=\dfrac{1}{\sqrt{2}+\sqrt{5}+1}=\dfrac{\sqrt{2}-\sqrt{5}+1}{(\sqrt{2}+\sqrt{5}+1)(\sqrt{2}-\sqrt{5}+1)}=\dfrac{\sqrt{2}-\sqrt{5}+1}{-2+2\sqrt{2}} A = ( 2 − 5 + 1 ) ( ( 1 + 2 ) ( − 2 + 2 2 ) ( 1 + 2 ) = 2 2 − 5 + 3 − 10 2 A=\dfrac{(\sqrt{2}-\sqrt{5}+1)((1+\sqrt{2})}{(-2+2\sqrt{2})(1+\sqrt{2})}=\dfrac{2\sqrt{2}-\sqrt{5}+3-\sqrt{10}}{2} A = 2 2 − 5 + 3 − 10 2 − 2 + 2 A=\dfrac{2\sqrt{2}-\sqrt{5}+3-\sqrt{10}}{2}-\sqrt{2}+\sqrt{2} A = 2 2 − 5 + 3 − 10 − 2 2 2 + 2 A=\dfrac{2\sqrt{2}-\sqrt{5}+3-\sqrt{10}-2\sqrt{2}}{2}+\sqrt{2} A = 3 − 5 − 10 2 + 2 A=\dfrac{3-\sqrt{5}-\sqrt{10}}{2}+\sqrt{2} A = 3 − 5 ( 2 + 1 ) 2 + 2 A=\boxed{\dfrac{3-\sqrt{5}(\sqrt{2}+1)}{2}+\sqrt{2}} and 3 − 5 + 2 = 0 3-5+2=\boxed{0}

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complain= it could have been written as 3 − 10 ( 1 2 + 1 ) 2 + 2 \dfrac{3-\sqrt{10}(\sqrt{\dfrac{1}{2}}+1)}{2}+\sqrt{2} and many more. please specify @Kartik Sharma

Aareyan Manzoor - 6 years, 5 months ago

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@Kartik Sharma Also remove the "-" sign from "can be written as - ", quite misleading...

Satvik Golechha - 6 years, 5 months ago

But you have forgotten that you should not have a radical sign on the denominator. I was refering to 1 2 \sqrt{\frac{1}{2}} .

Angelo Marco Ramoso - 6 years, 5 months ago
Ronak Agarwal
Jan 4, 2015

So all you wanted is 0 as a answer.

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