1. ## Extra roots?

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)

2. $\displaystyle \sqrt{x^2+1}=89-x^2$
Before squaring both sides and solve for x, you must put the condition that the right-hand member is nonnegative.
$\displaystyle 89-x^2\geq 0\Rightarrow x\in[-\sqrt{89},\sqrt{89}]$.
So, the other two roots, $\displaystyle \pm 3\sqrt{11}$ are not in this interval.

Another way to solve the equation:
Let $\displaystyle \sqrt{x^2+1}=t, \ t>0$.
Then $\displaystyle t^2+t-90=0\Rightarrow t_1=-10, \ t_2=9$
But $\displaystyle t>0\Rightarrow t=9\Rightarrow x^2+1=81\Rightarrow x=\pm 4\sqrt{5}$.

3. Thanks, I guess that's something new I'll have to watch out for