All About 2018 -- E

Algebra Level 3

x + 2 x + 2 x + . . . + 2 x + 2 3 x ⏟ Number of x = 2018 = x \underbrace{\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2\sqrt{3x}}}}}}_\text{Number of x = 2018}=x

Find the sum of all the real roots of the equation above.

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The answer is 3.

Harsh Shrivastava by Harsh Shrivastava , 17, India

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1 solution

X X
Jun 7, 2018

Let 3 x = x \sqrt{3x}=x .

Then x = x + 2 x + 2 x + . . . + 2 x + 2 3 x x=\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2\sqrt{3x}}}}}

= x + 2 x + 2 x + . . . + 2 x + 2 x =\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}

= x + 2 x + 2 x + . . . + 2 x = . . . = 3 x =\sqrt{x+2\sqrt{x+2\sqrt{x+...+2x}}}=...=\sqrt{3x}

It fits! So x = 3 x=3

1 Helpful 0 Interesting 0 Brilliant 0 Confused

How does this justify that no other solution exists?

typical beam - 2 years, 9 months ago

If x<3 then, x+ 2 root 3x > x + 2 root x^2 = 3x > x^2

so x+ 2 root { x+ 2 root 3x } > x +2 root { x^2 } = 3x > x^2

as same way,

x+ 2 root { x+ 2 root { ... + 2 root { x+ 2 root 3x } } } > x^2

so x > x : contradiction

As same way : If x >3 : contradiction

So 3 is unique solution

Dong kwan Yoo - 2 years, 2 months ago

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