An algebra problem by Mateus Gomes

Algebra Level 4

$\sqrt{x} + \frac1{\sqrt{x}} = 3$

$x - \dfrac1{x}= \sqrt?$

1 solution

A Former Brilliant Member
Jan 30, 2016

$\Rightarrow \sqrt{\color{#D61F06}{x}}+\frac{1}{\sqrt{\color{#D61F06}{x}}}=3$

Squaring both sides.

$\color{#D61F06}{x}+\frac{1}{\color{#D61F06}{x}}+\color{#EC7300}{2}=9$

$\color{#D61F06}{x}+\frac{1}{\color{#D61F06}{x}}=7$

Again, Squaring both sides.

$\color{#D61F06}{x}^2+\frac{1}{\color{#D61F06}{x}^2}+\color{#EC7300}{2}=49$

$\color{#D61F06}{x}^2+\frac{1}{\color{#D61F06}{x}^2}=47$

$\left(\color{#D61F06}{x}-\frac{1}{\color{#D61F06}{x}}\right)^2+\color{#EC7300}{2}=47$

$\left(\color{#D61F06}{x}-\frac{1}{\color{#D61F06}{x}}\right)=\sqrt{45}$

$\sqrt{45}=\sqrt{\color{#3D99F6}{a}}$

$\therefore \color{#3D99F6}{a}=\color{#20A900}{\boxed{45}}$

Shouldn't it be a level 3 problem !?

Venkata Karthik Bandaru - 5 years, 4 months ago

Yes, it would be if it was not reported.Earlier this question was wrong.

A Former Brilliant Member - 5 years, 4 months ago

Nope, level 2 is much.

Swapnil Das - 5 years, 4 months ago

Why was this Level 4?!

Mehul Arora - 5 years, 3 months ago