<s>[INST] Solution of given equation for $x$ Solve the given equation for $x$
$$\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3}=125$$
I solved the question by taking ${x^2-2x+3}=t$, and squaring twice and finally solving ${x^2-2x+3}=t$ but it required very hectic calculations. I wonder if someone can suggest better approach.
 [/INST] As $8-3=x^2-2x+8-(x^2-2x+3)$
$=(\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3})(\sqrt{x^2-2x+8}-\sqrt{x^2-2x+3})$
$$\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3}=125\iff(\sqrt{x^2-2x+8}-\sqrt{x^2-2x+3})=\dfrac{8-3}{125}$$
Adding we get $2\sqrt{x^2-2x+8}=125+\dfrac1{25}=?$
Now square both sides. 
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