How to solve an irrational equation?

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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 $5x^3-8x^2+65x-50 = 0$.