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Surd Equation Challenge

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
Solve \(\displaystyle \sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}\).
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

\(\displaystyle x=\phi=\frac{1+\sqrt{5}}{2}\)
 
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  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
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!