The problem is solve for x:

$\displaystyle \sqrt{x^2+1}+x^2+1=90$

If I do this in the 'normal' way, by isolating the square root and squaring both sides, I get:

$\displaystyle x^4-179x^2+7920=0$

$\displaystyle (x^2-80)(x^2-99)=0$

$\displaystyle x^2 = 80, 99$

This yields $\displaystyle x = \pm 4\sqrt{5}, \pm 3\sqrt{11}$

Why did the two extraneous roots $\displaystyle \pm 3\sqrt{11}$ appear?

Thanks!

(I realised you can immediately set $\displaystyle y=\sqrt{x^2+1}$, but that's not the point)