Solution of given equation for $x$

Question

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.

Answer 1

Like Will Jagy said, substitute $t=\sqrt{x^2 -2x+3}$. You then can solve the formula: $$\sqrt{t^2+5}=125-t \\t^2 + 5 =15625-250t +t^2$$ $$250t = 15620$$ $$t=62.48$$ Then you can plug your value for $t$ back into $t^2=x^2 -2x+3$ and solve a quadratic formula with respect to $x$.

Answer 2

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.