5
$\begingroup$

Solve for $x \in \mathbb{R}$

$$\dfrac{\sqrt{x^2-x+2}}{1+\sqrt{-x^2+x+2}} - \dfrac{\sqrt{x^2+x}}{1+\sqrt{-x^2-x+4}} = x^2-1$$

I tried squaring the equation but it became a sixteen degree equation. I also tried substitutions, but that didn't help. There must be some elegant solution to it in its current form.

Any help will be appreciated.
Thank you.

$\endgroup$
  • 1
    $\begingroup$ It's easy to see that $x = 1$ is a solution (I'm not sure what the others are, if any). $\endgroup$ – Irregular User May 4 '16 at 0:05
  • $\begingroup$ @IrregularUser Is it the only solution? Can you prove it? $\endgroup$ – Henry May 4 '16 at 0:06
  • $\begingroup$ I can't prove or disprove whether it's the only solution or not, but I'm sure that one of the comments in the future will answer your question. $\endgroup$ – Irregular User May 4 '16 at 0:07
  • 1
    $\begingroup$ @Moo I'm seeking a mathematical solution using algebra i.e, substitutions etc. $\endgroup$ – Henry May 4 '16 at 0:15
  • $\begingroup$ I'm wondering if multiplying both sides by $(1+\sqrt{-x^2+x+2})(1+\sqrt{-x^2-x+4})$ would help by clearing the denominator... Just a thought $\endgroup$ – Frank May 4 '16 at 0:33
8
$\begingroup$

Firstly, we need to find what values of $x$ are acceptable. Solving the inequalities: $$\begin{array}[t]{rl} x^2-x+2 &\ge 0\\ x^2+x &\ge 0 \\ -x^2+x+2 &\ge 0\\ -x^2-x+4 &\ge 0 \end{array} $$ yields $$\left.\begin{array}{l} x \in \mathbb R\\ x \in (-\infty,-1] \cup [0,+\infty)\\ x \in [-1,2] \\ x\in \left[ \frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2}\right] \end{array}\right\}\implies x\in \{-1\} \cup \left[0,\frac{-1+\sqrt{17}}{2}\right]. $$

We can see that $x = -1$ is not a solution, so we are looking for non - negative solutions.


For $ x \in [0,1)$ the RHS is negative. But in that case we have that the first numerator is greater than the second one (this can be derived easily by solving the inequality $\sqrt{x^2-x+2} \gt \sqrt{x^2+x}$) and the first denominator is less than the second denominator, thus the first fraction is always greater than the second one, so their difference is positive. Thus there are no solutions in $[0,1)$.

Similarly, for $x\in \left(1,\frac{-1+\sqrt{17}}{2}\right]$ we have that the RHS is positive and the LHS is negative.

This implies that the only real solution is $x = 1$.

$\endgroup$
  • $\begingroup$ I get your first part, but when you say "similarly for $x\in \left(1,\frac{-1+\sqrt{17}}{2}\right]$ ", I don't think it will be easy to establish this claim. $\endgroup$ – Henry May 4 '16 at 1:31
  • 2
    $\begingroup$ Solving the inequality $\sqrt{x^2 - x+ 2} \lt \sqrt{x^2 + x}\implies x \gt 1$ and solving the inequality $1+ \sqrt{-x^2+x+2}\gt 1+\sqrt{-x^2-x+4}\implies x>1.$ Thus, for $x > 1$ we have that the first fraction is less than the second one. But $x$ must be less than $\frac{-1+\sqrt{17}}{2},$ that's why I mentioned for $x \in \left(1,\frac{-1+\sqrt{17}}{2}\right]$ blah blah blah... $\endgroup$ – thanasissdr May 4 '16 at 1:40
  • $\begingroup$ Ah right. Thanks a lot :) $\endgroup$ – Henry May 4 '16 at 1:44
  • $\begingroup$ You are welcome! :) $\endgroup$ – thanasissdr May 4 '16 at 1:45
  • $\begingroup$ Hello Sir. Can you also help me with this question ? $\endgroup$ – Henry May 4 '16 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.