# How do I solve this equation

How do I solve this equation, I keep getting the wrong answer.

$$\frac{-2x^2+50}{\sqrt{50-x^2}} = 1$$

Thank you

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$$\dfrac{-2x^2+50}{\sqrt{50-x^2}} = 1 \quad \Rightarrow \quad -2x^2+50=\sqrt{50-x^2}$$ Substitute $y=x^2\ge 0$, square both sides and you get a degree $2$ equation to solve for positive values of $y$.

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If you put $y=x^2$ you will see that you get a quadratic in y. Squaring like this can add solutions (the ones which belong to the negative square root) - so you have to check back to make sure you identify the solutions to the original problem out of the solutions you get after squaring. – Mark Bennet Oct 30 '12 at 15:12
@MarkBennet thanks, I'll edit it. – user31280 Oct 30 '12 at 15:14

First clear the fraction to get $-2x^2+50=\sqrt{50-x^2}$. Now square both sides to get rid of the square root, but be careful: squaring can introduce extraneous solutions. That is, the new equation may have more solutions than the original one. Thus, you must be sure to check all of your solutions, throwing out any extraneous ones.

Squaring gives you $4x^4-200x^2+2500=50-x^2$. Collect everything on one side:

$$4x^4-199x^2+2450=0\;.\tag{1}$$

This looks a bit ugly, since we have a fourth degree polynomial, but it’s not really bad: you have only even powers of $x$, and this is a disguised quadratic in the variable $x^2$. If you’ve not encountered this kind of equation before, you may find it helpful to make the substitution $y=x^2$ explicitly, so that $(1)$ becomes $4y^2-199y+2450=0$, an ordinary quadratic. Solve the quadratic for $y$, and you’ll have $x^2$, from which you can easily find $x$.

Alternatively, since only $x^2$ appears in the original equation, you could substitute $y=x^2$ in that to get $-2y+50=\sqrt{50-y}$, square to get $4y^2-200y+2500=50-y$, and proceed as before.

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The solutions given above are the reasonable ones. Here is a variant. We look only for real solutions. Let $50-x^2=u^2$ where without loss of generality we may take $u\gt 0$.

Then $50-2x^2=50-2(50-u^2)=2u^2-50$. So our equation becomes $$\frac{2u^2-50}{u}=1,$$ or equivalently $2u^2-u-50=0$. The positive solution of this is $u=\dfrac{1+\sqrt{401}}{4}$. It follows that $$x=\pm \sqrt{50-\left( \frac{1+\sqrt{401}}{4} \right)^2}.$$

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We have:

$$\sqrt{50 - x^2} = 50 - 2x^2$$

Eliminating the square root and opening the brackets:

$$50 - x^2 = 4x^4 - 200x^2 + 2500$$

Make it more simple:

$$-4x^4 + 199x^2 - 2450 = 0$$

Substitute $y = x^2$, simplifying and find full square:

$$\left(y - \dfrac{199}{8}\right)^2 = \dfrac{401}{64}$$

So, there is two solutions:

$$y = \dfrac{199}{8} \pm \dfrac{\sqrt{401}}{8}$$

So, there is 4 solutions for $x$:

$$x = \pm\sqrt{\dfrac{199}{8} - \dfrac{\sqrt{401}}{8}}\tag{1}$$ $$x = \pm\sqrt{\dfrac{199}{8} + \dfrac{\sqrt{401}}{8}}\tag{2}$$

Substitution the solution for $x$ into original equations shows the correcteness of $(1)$ solutions and incorrectness of solutions $(2)$.

Finally: $$x_1 = \sqrt{\dfrac{199}{8} - \dfrac{\sqrt{401}}{8}}\approx 4.7299\qquad x_2 = -\sqrt{\dfrac{199}{8} - \dfrac{\sqrt{401}}{8}}\approx -4.7299$$

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