Radicals are complicated(2)

Algebra Level 5

1 2016 x − 2016 x − 1 2016 x + 2016 x = 1 2016 x − 1 \frac{1}{\sqrt{2016x-\sqrt{2016x}}}-\frac{1}{\sqrt{2016x+\sqrt{2016x}}} = \frac{1}{\sqrt{2016x-1}}

If the sum of all real root(s) of the above equation is of the form 1 y \large \frac{1}{y} .

Find y y .


The answer is 1512.

Tommy Li by Tommy Li , 22, Hong Kong

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1 solution

Tommy Li
Jun 15, 2016

1 2016 x − 2016 x − 1 2016 x + 2016 x = 1 2016 x − 1 \large \frac{1}{\sqrt{2016x-\sqrt{2016x}}}-\frac{1}{\sqrt{2016x+\sqrt{2016x}}} = \frac{1}{\sqrt{2016x-1}}

2016 x + 2016 x − 2016 x − 2016 x ( 2016 x ) 2 − 2016 x = 1 2016 x − 1 \large \frac{\sqrt{2016x+\sqrt{2016x}}-\sqrt{2016x-\sqrt{2016x}}}{\sqrt{(2016x)^{2}-2016x}} = \frac{1}{\sqrt{2016x-1}}

2016 x + 2016 x − 2016 x − 2016 x = ( 2016 x ) 2 − 2016 x 2016 x − 1 \sqrt{2016x+\sqrt{2016x}}-\sqrt{2016x-\sqrt{2016x}} =\large \frac{\sqrt{(2016x)^{2}-2016x}}{\sqrt{2016x-1}}

2016 x + 2016 x − 2016 x − 2016 x = 2016 x \sqrt{2016x+\sqrt{2016x}}-\sqrt{2016x-\sqrt{2016x}} = \sqrt{2016x}

( 2016 x + 2016 x − 2016 x − 2016 x ) 2 = ( 2016 x ) 2 (\sqrt{2016x+\sqrt{2016x}}-\sqrt{2016x-\sqrt{2016x}})^{2} = (\sqrt{2016x})^{2}

2016 x + 2016 x − 2 ( 2016 x + 2016 x ) ( 2016 x − 2016 x ) + 2016 x − 2016 x = 2016 x 2016x+\sqrt{2016x} - 2(\sqrt{2016x+\sqrt{2016x}})(\sqrt{2016x-\sqrt{2016x}}) +2016x-\sqrt{2016x}= 2016x

2016 x = 2 ( 2016 x ) 2 − 2016 x 2016x=2\sqrt{(2016x)^{2}-2016x}

( 2016 x ) 2 = ( 2 ( 2016 x ) 2 − 2016 x ) 2 (2016x)^2=(2\sqrt{(2016x)^{2}-2016x})^2

201 6 2 x 2 = 4 ( 201 6 2 x 2 − 2016 x ) 2016^2x^2=4(2016^2x^2-2016x)

3 × 201 6 2 x 2 − 4 × 2016 x = 0 3 \times 2016^2x^2- 4 \times 2016x = 0

4 x ( 1512 x − 1 ) = 0 4x(1512x-1)=0

x = 0 x = 0 (rej.) or x = 1 1512 x = \frac{1}{1512}

y = 1512 y=1512

2 Helpful 0 Interesting 0 Brilliant 0 Confused

L.H.S = 1 2016 ( 1 1512 ) − 2016 ( 1 1512 ) − 1 2016 ( 1 1512 ) + 2016 ( 1 1512 ) \large = \frac{1}{\sqrt{2016 (\frac{1}{1512})-\sqrt{2016 (\frac{1}{1512})}}}-\frac{1}{\sqrt{2016 (\frac{1}{1512})+\sqrt{2016( \frac{1}{1512})}}}

= 1 4 3 − 4 3 − 1 4 3 + 4 3 \large = \frac{1}{\sqrt{\frac{4}{3}-\sqrt{\frac{4}{3}}}}-\frac{1}{\sqrt{\frac{4}{3}+\sqrt{\frac{4}{3}}}}

= 1 4 − 2 3 3 − 1 4 + 2 3 3 \large = \frac{1}{\sqrt{\frac{4-2\sqrt{3}}{3}}}-\frac{1}{\sqrt{\frac{4+2\sqrt{3}}{3}}}

= 1 3 − 1 3 − 1 3 + 1 3 \large = \frac{1}{\frac{\sqrt{3}-1}{\sqrt{3}}}-\frac{1}{\frac{\sqrt{3}+1}{\sqrt{3}}}

= 3 ( ( 3 + 1 ) − ( 3 − 1 ) ( 3 − 1 ) ( 3 + 1 ) ) = \sqrt{3}(\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}-1)(\sqrt{3}+1)})

= 3 = \sqrt{3}

R.H.S = 1 2016 ( 1 1512 ) − 1 = \frac{1}{\sqrt{2016(\frac{1}{1512})-1}}

= 1 4 3 − 1 = \frac{1}{\sqrt{\frac{4}{3}-1}}

= 1 ( 1 3 ) = \frac{1}{(\frac{1}{\sqrt{3}})}

= 3 = \sqrt{3}

L.H.S = R.H.S

Tommy Li - 5 years ago

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