# How to solve an irrational equation?

I want to solve this equation $$2 (x-2) \sqrt{5-x^2}+(x+1)\sqrt{5+x^2} = 7 x-5.$$ I tried The given equation equavalent to $$2 (x-2) (\sqrt{5-x^2}-2)+(x+1)(\sqrt{5+x^2}- 3)=0$$ or $$(x-2)(x+1)\left [\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2}\right ] = 0.$$ I see that, the equation $$\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2} = 0$$ has unique solution $x = 2$, but I can not solve. How can I solve this equation or solve the given equation with another way?

• Try isolating one of the square roots on one side of the equation, squaring both sides, then isolating the remaining square root on one side of the equation and squaring both sides again. This will eliminate the square roots. Dec 27, 2015 at 2:14

Squaring the equation, we obtain \begin{align} 4(x-2)^2(5-x^2) + (x+1)^2(5+x^2) + 4(x-2)(x+1)\sqrt{25-x^4} & = (7x-5)^2 \end{align} Simplifying the above, we obtain that \begin{align} 4(x-2)(x+1)\sqrt{25-x^4} & = 3x^4-18x^3+39x^2-60 = 3(x-2)(x+1)(x^2-5x+10) \end{align} This gives us either $x=2$ or $x=-1$ or $$4\sqrt{25-x^4} = 3(x^2-5x+10) = 3\left(\left(x-\dfrac52\right)^2 + \dfrac{15}4\right) = \dfrac{45}4 + 3\left(x-\dfrac52\right)^2$$ Plugging in $x=2$ or $x=-1$ in the original equation, we see that $x=2$ or $x=-1$ are valid solutions.

The only other possibility is when $$16\sqrt{25-x^4} = 45 + 3\left(2x-5\right)^2 = 12x^2-60x+120 \implies 4\sqrt{25-x^4} = 3x^2-15x+30$$ The only integer solution we can hope is when $x$ is an integer and $25-x^4$ is a square, which gives us that $x=2$. Squaring both sides, we obtain \begin{align} 16\left(25-x^4\right) & = \left(3x^2-15x+30\right)^2\\ 400-16x^4 & = 9x^4 + 225x^2 + 900 -90x^3 + 180x^2 - 900x\\ 25x^4 - 90x^3+405x^2-900x+500 & = 0\\ 5x^4 - 18x^3 + 81x^2 - 180x + 100 & = 0 \end{align} As we saw earlier $x=2$ should be a solution to this. Hence, we have $$5x^4 - 18x^3 + 81x^2 - 180x + 100 = (x-2)\left(5x^3-8x^2+65x-50\right)$$ Hence, the only other possible root should satisfy $5x^3-8x^2+65x-50 = 0$.

• There is a solution to the final equation, around 0.8, but it doesn't satisfy the original solution. Dec 27, 2015 at 7:46

First of all note that the RHS of your original equation can be written as $4(x-2)+3(x+1)$. Now transfer the terms on either side of the equation obtaining $$2(x-2)\big[\sqrt {5-x^2}-2\big] = (x+1)\big[3-\sqrt{5+x^2}\big].$$ The LHS vanishes for $\pm 1$ and $2$. The RHS vanishes for $-1$ and $\pm2$. Two of the root are therefore $-1$ and $2$.

EDIT: Since the RHS remains positive and the LHS remains negative from $-1$ to $2$, there are no further roots between these two. In the range $-\sqrt 5$ to -1, the LHS is more than RHS and in the range $2$ to $\sqrt 5$, LHS is less than RHS

• Why does the common value need to be zero in your equation? Your roots are correct per Alpha You have shown that $-1$ and $2$ are roots but have not shown that there are not more. Dec 27, 2015 at 3:12
• Yes it remains to be shown that the two sides do not agree on other values of x. I am checking it on a plot. Will update in a while.
– vnd
Dec 27, 2015 at 3:15

The usual way to do something like this is to square to get rid of one radical, rearrange, and square again to get rid of the remaining radical, as @vhspdfg says in a comment. I won’t detail the intermediate steps, but the octic polynomial I got was $$2000 - 1600x - 3480x^2 + 2960x^3 + 825x^4 - 1340x^5 + 510x^6 - 140x^7 + 25x^8\\ =5(x+1)^2(x-2)^3(5x^3 - 8x^2 + 65x - 50)\,.$$ The cubic seems to have no rational roots, and if this is correct, it’s irreducible.

• Later: there is only one real root of the cubic, around $0.8$, and as @DanielV has pointed out, it’s not a root of the original equation. Dec 27, 2015 at 16:07