Solving Tricky Algebraic Equation: Tips and Attempt at Solution

[SithsNGiggles]

Homework Statement

I'm helping a friend out with an optimization problem, and I wound up with a tricky equation in which I have to solve for x. Is there a particular way to go about doing this?

$4x^2 - 2x \sqrt{r^2 - x^2} - 2r^2 = 0$,

where $r$ is a positive constant.

Homework Equations

The Attempt at a Solution

So far I've only typed the equation into WolframAlpha, but the solution they provide is beyond my understanding.

 

[Gravitational]

Do you want to solve for x in terms of r?

 

[Bread18]

Put the sqrt on the other side of the eqn and square both sides, should be able to do it from there.

 

[SithsNGiggles]

Do you want to solve for x in terms of r?

Yes.

Put the sqrt on the other side of the eqn and square both sides, should be able to do it from there.

Oh, I don't know why I hadn't considered that. Unfortunately, after simplifying it a bit, I'm still not sure how to solve for x:
$4x^2 - 2x\sqrt{r^2 - x^2} - 2r^2 = 0$
$2x^2 - x\sqrt{r^2 - x^2} - r^2 = 0$
$2x^2 - r^2 = x\sqrt{r^2 - x^2}$
$(2x^2 - r^2)^2 = (x\sqrt{r^2 - x^2})^2$
$4x^4 - 4x^2r^2 + r^4 = x^2 (r^2 - x^2)$
$4x^4 - 4x^2r^2 + r^4 = x^2r^2 - x^4$
$5x^4 - 5x^2r^2 + r^4 = 0$

 

[Curious3141]

Homework Statement

I'm helping a friend out with an optimization problem, and I wound up with a tricky equation in which I have to solve for x. Is there a particular way to go about doing this?

$4x^2 - 2x \sqrt{r^2 - x^2} - 2r^2 = 0$,

where $r$ is a positive constant.

Homework Equations

The Attempt at a Solution

So far I've only typed the equation into WolframAlpha, but the solution they provide is beyond my understanding.

Bring the term with the square root to the RHS. Square both sides. You should now have a quartic in x, which is actually a quadratic in x². Solve for x². At this stage, put the values of x² you get back into the original equation to ensure they satisfy the original problem. Squaring equations may introduce redundant roots.

Finally, take square roots to get x.

 

[Curious3141]

Yes.

Oh, I don't know why I hadn't considered that. Unfortunately, after simplifying it a bit, I'm still not sure how to solve for x:
$4x^2 - 2x\sqrt{r^2 - x^2} - 2r^2 = 0$
$2x^2 - x\sqrt{r^2 - x^2} - r^2 = 0$
$2x^2 - r^2 = x\sqrt{r^2 - x^2}$
$(2x^2 - r^2)^2 = (x\sqrt{r^2 - x^2})^2$
$4x^4 - 4x^2r^2 + r^4 = x^2 (r^2 - x^2)$
$4x^4 - 4x^2r^2 + r^4 = x^2r^2 - x^4$
$5x^4 - 5x^2r^2 + r^4 = 0$

As I wrote, quartic in x, but a quadratic in x². To make it clearer, let x² = y and solve for y.

 

[micromass]

Put $y=x^2$ then you get a quadratic equation.

 

[Dick]

Yes.

Oh, I don't know why I hadn't considered that. Unfortunately, after simplifying it a bit, I'm still not sure how to solve for x:
$4x^2 - 2x\sqrt{r^2 - x^2} - 2r^2 = 0$
$2x^2 - x\sqrt{r^2 - x^2} - r^2 = 0$
$2x^2 - r^2 = x\sqrt{r^2 - x^2}$
$(2x^2 - r^2)^2 = (x\sqrt{r^2 - x^2})^2$
$4x^4 - 4x^2r^2 + r^4 = x^2 (r^2 - x^2)$
$4x^4 - 4x^2r^2 + r^4 = x^2r^2 - x^4$
$5x^4 - 5x^2r^2 + r^4 = 0$

Now substitute u=x^2 and you have a quadratic equation in u. You can solve that, right?

 

[SithsNGiggles]

Ah, thanks. Making that kind of substitution isn't as second-nature as it should be.