Surd Equation Challenge

anemone

MHB POTW Director
Staff member
Solve $$\displaystyle \sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}$$.

MarkFL

Administrator
Staff member
My solution:

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

Arranging as:

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

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

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

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

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

From which, we obtain the only valid root is:

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

anemone

MHB POTW Director
Staff member
My solution:

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

Arranging as:

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

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

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

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

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

From which, we obtain the only valid root is:

$$\displaystyle x=\phi=\frac{1+\sqrt{5}}{2}$$
Hi MarkFL,

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