How to Solve the Surd Equation Challenge \sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}?

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The discussion focuses on solving the surd equation $$\sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}$$. MarkFL provided a correct solution, demonstrating a clear and effective approach to the problem. The community acknowledged the validity of the solution, highlighting the importance of methodical reasoning in tackling such equations.

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Solve $$\sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}$$.
 
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My solution:

We see that we require $$1\le x$$.

Arranging as:

$$\sqrt{x^2-1}=x\sqrt{x}-\sqrt{x-1}$$

Squaring, adding through by $1$, then dividing through by $x\ne0$, we obtain:

$$2\sqrt{x(x-1)}=x^2-x+1$$

Squaring again, collecting like terms, and factoring, we obtain:

$$\left(x^2-x-1 \right)^2=0$$

From which, we obtain the only valid root is:

$$x=\phi=\frac{1+\sqrt{5}}{2}$$
 
MarkFL said:
My solution:

We see that we require $$1\le x$$.

Arranging as:

$$\sqrt{x^2-1}=x\sqrt{x}-\sqrt{x-1}$$

Squaring, adding through by $1$, then dividing through by $x\ne0$, we obtain:

$$2\sqrt{x(x-1)}=x^2-x+1$$

Squaring again, collecting like terms, and factoring, we obtain:

$$\left(x^2-x-1 \right)^2=0$$

From which, we obtain the only valid root is:

$$x=\phi=\frac{1+\sqrt{5}}{2}$$

Hi MarkFL,

Thanks for participating! Your answer is correct and the way you approached it is brilliant!
 

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