# Solve the equation $\sqrt{x+5}=5-x^2$

Solve the equation $\sqrt{x+5}=5-x^2$. I have tried to make the substitution $x=\sqrt{5}\tan^2 \theta$ and wanted to make use of the identity $\tan^2\theta+1=\sec^2\theta$ but it didn't work out. I also tried to make the substitution $y=x+5$ but it lead to nowhere. Since this was a contest problem, I believe there is a short, elegant and elementary solution, please helps.

• Did you try the substitution $x=5\tan^2 \theta$? – Brian Cheung Sep 4 '15 at 10:59
• Put $x = 5\tan^2 \theta$. Then, we have $$\sqrt{\sec^2 \theta} = 5(1-\tan^2 \theta)\sec^2\theta$$ – GAVD Sep 4 '15 at 11:02
• OMG I got it, thanks so much guys! – nayr ktn Sep 4 '15 at 11:04
• Why not squaring both sides and factoring the resultng equation ? – Peter Sep 4 '15 at 11:06
• @Peter, I tried but I couldn't factorize it. – nayr ktn Sep 4 '15 at 11:09

# Hint:

the equation equivalen to $$x^4-10x^2-x+20=0$$ Because there is no $x^3$ in the equation and the coefficient of $x^4$ is $1$ , so we can use the following $$(x^2-x+A)(x^2+x+B)=x^4-10x^2-x+20$$

Here $5+x\geq 0\Rightarrow x\geq -5$ and $5-x^2\geq 0\Rightarrow x^2-\left(\sqrt{5}\right)^2\leq 0\Rightarrow -\sqrt{5}\leq x \leq \sqrt{5}$

So we get $-\sqrt{5}\leq x \leq \sqrt{5}.$

Now Let $\sqrt{5+x}=y\;,$ Then $$y^2=5+x \tag{1}$$

and equation convert into $$y=5-x^2\Rightarrow x^2=5-y\tag{2}$$

So $$y^2-x^2 = 5+x-(5-y)=(y+x)\Rightarrow (y^2-x^2)=(y+x)$$

So $$(y+x)\cdot(y-x)-(y+x) =0\Rightarrow (y+x)\cdot \left[y-x-1\right]=0$$

So either $y=x$ or $y=x+1$

$\bullet \;$ If $y=x\;,$ Then put into $y^2=5+x\Rightarrow x^2=5+x$

So we get $$\displaystyle x^2-x-5=0\Rightarrow x=\frac{1\pm \sqrt{1+20}}{2}=\frac{1\pm\sqrt{21}}{2}$$

So we get $$\displaystyle x=\frac{1-\sqrt{21}}{2}.$$ bcz here $-\sqrt{5}\leq x\leq \sqrt{5}$

$\bullet \;$ If $y=1+x\;,$ Then put into $y^2=5+x\Rightarrow (1+x)^2=5+x$

So we get $$1+x^2+2x=5+x\Rightarrow \displaystyle x^2+x-4=0\Rightarrow x= \frac{-1\pm\sqrt{17}}{2}$$

So we get $$\displaystyle x= \frac{-1+\sqrt{17}}{2}$$ bcz here $-\sqrt{5}\leq x\leq \sqrt{5}$

So final solution is $\displaystyle x = \left\{\displaystyle \frac{-1+\sqrt{17}}{2}\;, \frac{1-\sqrt{21}}{2} \right\}$