Not just an equation,there's more!

Algebra Level 4

Solve for $x \in \mathbb{R}$ such that $\dfrac {|x^2 - 1|}{x - 2} = x.$ Convert the only real solution to a continued fraction of the form $-[0;\overline{a,b}]$ .Find $a+b$ .

The answer is 3.

1 solution

A Former Brilliant Member
Jan 10, 2016

We can split this question into two cases:

Case 1: $x^{2}-1\geq0\Rightarrow x\leq-1$ or $x\geq1$

Then, we have $x^{2}-1=x^{2}-2x\Rightarrow x=\frac{1}{2}$ , a contradiction.

Case 2: $x^{2}-1\leq0\Rightarrow -1\leq x\leq 1$

Then, we have $x^{2}-1=2x-x^{2}\Rightarrow 2x^{2}-2x-1=0$ .

Solving, we get $x=\frac{1}{2}(1-\sqrt{3})$ or $x=\frac{1}{2}(1+\sqrt{3})$ .

However, only the first solution satisfies the conditions, so it is the only real solution.

Thus, the continued fraction is easily obtained and therefore, $a+b=3$ .

Nice, same way(+1)

Shaun Leong - 5 years, 5 months ago

Why 1/2 is not a real solution?

Department 8 - 5 years, 5 months ago

Because the constraints we have taken are x>=1 or x <=-1 and x=0.5 doesnt fall in this interval.

Adarsh Kumar - 5 years, 5 months ago

Thanks didn't notice that

Department 8 - 5 years, 5 months ago

But the solution is equal to -0,3660254037... so where does this continued fraction come from?

Jan Chomiak - 4 years, 2 months ago

Not just an equation,there's more!

The answer is 3.

1 solution

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