Find all real values of $x$ which satisfy $\sqrt{x^2 + 1} + x^2 + 1 = 90$.

Question

Find all real values of $x$ which satisfy $\sqrt{x^2 + 1} + x^2 + 1 = 90$.

\begin{align*} \sqrt{x^2+1} + x^2 &= 89 \\ {(\sqrt{x^2+1} + x^2)}^2 &= 89^2 \\ x^2+2x^2\sqrt{x^2+1}+x^4 &=89^2 \end{align*}

Letting $y = x^2$, so $y^2 = x^4$. This gives us \begin{equation} y^2+2y\sqrt{y+1}+y=7920 \end{equation}

I not sure how to proceed to eliminate the radical. I shouldn't square quantities that aren't both positive or else I'll lose the equivalency in the new equation right?

Answer 1

Let me point out couple of substitutions that simplifies this equation:

1. $T=\sqrt{x^2+1},$ then we have a quadratic equation $$T^2+T-90=0$$ which has two solution s $T=9$ and $T=-10.$ Then you continue from here (with the condition that $T\ge 1$).

2. $x=\tan\theta,$ then $x^2+1=\sec^2\theta$ and we get a trigonometric equation $$\sec^2\theta+\sec\theta-90=0.$$ Similarly $x=\cot\theta$ is also a possible trigonometric substitution for this equation.

3. $x=\sinh\theta$ has the property that $x^2+1=\cosh^2\theta$ and hence we get $$\cosh^2\theta+\cosh\theta-90=0.$$

In last two substitutions both $\sec$ and $\cosh$ are even functions, therefore we don't need to consider the negative square root as a separate case.

Answer 2

Hint

Begin by putting $$y=\sqrt{x^2+1}$$

with

$$y+y^2=90$$ or $$y^2+y-90=0$$ $$y=\frac{-1\pm 19}{2}$$ and $ y>0$.

You should find that $$x=\pm \sqrt{80}$$

Answer 3

Hint to solve

$x^2+1=(89-x^2)^2$

Second hint

Turn the $x^4$ to a variable like $z=x^2$ so you can solve the exercise using discriminant.