What are the roots of this equation? (Quadratic Equation)

Question

$\frac{x}{\sqrt{1-x^2}}=A \Rightarrow \frac{x^2}{1-x^2}=A^2$

$\Rightarrow x^2=A^2(1-x^2)=A^2-A^2x^2$

$\Rightarrow x^2+A^2x^2=A^2$

$\Rightarrow x^2(1+A^2)=A^2$

$\Rightarrow x^2(1+A^2)-A^2=0$

I have tried this;

$\Delta=b^2-4ac=-4(1+A^2)(-A^2)=4(1+A^2)(A^2)$

$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{\pm2A\sqrt{1+A^2}}{{2(1+A^2)}}$

$x_{1,2}=\frac{\pm A\sqrt{1+A^2}}{1+A^2}$

But I got nothing.

Answer 1

Your algebra is fine. Another way of writing your answer (in a more simplified form) would be $$x=\pm\frac{A}{\sqrt{1+A^2}}$$ since $$\frac{\sqrt{t}}t=\frac1{\sqrt{t}}.$$

Answer 2

So $x_{1,2}=\frac{\pm A\sqrt{1+A^2}}{1+A^2}$.

That's it. Go home. Eat lunch. You are done.

Well, not entirely. $\sqrt{1-x^2}$ must not equal $0$ and $1-x^2\ge 0$. Need to check that this holds for $x_{1,2}$.

$\sqrt{1-x^2} = 0 \iff x^2 = 1\iff x=\pm 1$. And $1-x^2 \ge 0 \iff x^2 \le 1 \iff |x| \le 1$. So we must show $|x_{1,2}| < 1$.

$|A| = \sqrt {A^2} < \sqrt {1+A^2}$. So $|A|\sqrt{1+A^2} < \sqrt{1+A^2}^2 =1+A^2$

So $|x_{1,2}| = \frac {|A|\sqrt{1+A^2}}{1+A^2} < \frac {1+A^2}{1+A^2} = 1$.

Also you can simplify $x_{1,2} = \frac {\pm A}{\sqrt{1+A^2}}$ but that is not required.

Answer 3

As you seem to find $x = \frac {A\sqrt{A^2 + 1}}{A^2 + 1}$ to be unsatisfying.

I don't know if you find this any more satisfying.

$\frac {x}{\sqrt {1-x^2}} = \tan (\arcsin x)= A\\ x = \sin (\arctan A)$