Solve $\frac{\left(x-7\right)}{\sqrt{x-3}+2}+\frac{\left(x-5\right)}{\sqrt{x-4}+1}=\sqrt{10}$

Question

The answer in wolfram is 13. Any easier technique to solve this equation?

Answer 1

After the rationalization we get $$\sqrt{x-3}-2+\sqrt{x-4}-1=\sqrt{10}$$

so $$\sqrt{x-3}+\sqrt{x-4}=\sqrt{10}+3$$

Now the function on the left is strictly increasing so the equation has at most one solution and this is easy to guess...

Answer 2

$$\frac{x-7}{\sqrt{x-3}+2}+\frac{x-5}{\sqrt{x-4}+1}=\sqrt{10}$$ Substitute $x=y+4$ $$\frac{y-3}{\sqrt{y+1}+2}+\frac{y-1}{\sqrt{y}+1}=\sqrt{10}$$ $$\frac{y-3}{\sqrt{y+1}+2}+\frac{(\sqrt{y}+1)(\sqrt{y}-1)}{\sqrt{y}+1}=\sqrt{10}$$ $$\frac{y-3}{\sqrt{y+1}+2}+\sqrt{y}-1=\sqrt{10}$$ Substitute $y=z-1$ $$\frac{z-4}{\sqrt{z}+2}+\sqrt{z-1}-1=\sqrt{10}$$ $$\frac{(\sqrt{z}-2)(\sqrt{z}+2)}{\sqrt{z}+2}+\sqrt{z-1}-1=\sqrt{10}$$ $$\sqrt{z}-2+\sqrt{z-1}-1=\sqrt{10}$$ $$\sqrt{z-1}=3+\sqrt{10}-\sqrt{z}$$ $$(\sqrt{z-1})^2=\left(-\sqrt{z}+\sqrt{10}+3\right)^2$$ $$z-1=z-2 \sqrt{10 z}-6 \sqrt{z}+6 \sqrt{10}+19$$ $$z-\left(z-6 \sqrt{z}-2 \sqrt{10 z}+6 \sqrt{10}+19\right)-1=0$$ $$2 \sqrt{10} \sqrt{z}+6 \sqrt{z}-6 \sqrt{10}-20=0$$ Substitute $\sqrt{z}=w$ $$2 \sqrt{10} w+6 w-6 \sqrt{10}-20=0$$ $$\left(6+2 \sqrt{10}\right) w-6 \sqrt{10}-20=0$$ $$w=\frac{10+3 \sqrt{10}}{3+\sqrt{10}}\to w=\sqrt{10}$$ $z=w^2$ so $z=10$ and $y=z-1=9$ and $x=y+4\to \color{red}{x=13}$

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