solve this equation $$\sqrt{\sqrt{3}-\sqrt{\sqrt{3}+x}}=x$$
My try: since $$\sqrt{3}-x^2=\sqrt{\sqrt{3}+x}$$ then $$(x^2-\sqrt{3})^2=x+\sqrt{3}$$
Asked
Active
Viewed 165 times
5
-
1The [tag:differential-equations] is inappropriate. Please read a tags' description before applying it. – AlexR Dec 16 '13 at 09:38
2 Answers
10
Let $y = \sqrt{\sqrt3 + x}$, then the equation becomes: $\sqrt3 - y = x^2$, and $y^2 = \sqrt3 + x$.
So:
$$\begin{align*}
y^2 - x & = y + x^2 \\
\Rightarrow (y + x)(y - x - 1) & = 0
\end{align*}$$
And we can go from here.
-
1Please use MathJax for mathematical expressions. For a quick tutorial, see the site FAQ. – AlexR Dec 16 '13 at 09:42
0
You're on the right track. Square again, you'll get $x=0$ as a possible solution. Remembering that $x^3-x=x(x-1)(x+1)$ you should be able to factor the equation and get $x=-1$ as a possible solution. Now solve the remaining quadratic.
Michael Hoppe
- 18,103
- 3
- 32
- 49