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- Feb 14, 2012

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

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- Thread starter
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- #1

- Feb 14, 2012

- 3,683

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

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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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- Feb 14, 2012

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Hi

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:

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

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