Question

If \sqrt{1-\dfrac{1}{x} }+\sqrt{x-\dfrac{1}{x} } =x

Find the value of x

Collected in the board: Quadratic function

Steven Zheng posted 7 hours ago

Answer

\sqrt{1-\dfrac{1}{x} } =x-\sqrt{x-\dfrac{1}{x} }

Square both sides

1-\dfrac{1}{x} = x^2+ x-\dfrac{1}{x} -2x\sqrt{x-\dfrac{1}{x} }

x^2+x-1 = 2x\sqrt{x-\dfrac{1}{x} }

Now we get a quadratic equation in terms of \sqrt{x-\dfrac{1}{x} }

x-\dfrac{1}{x} - 2\sqrt{x-\dfrac{1}{x} }+1=0

Then

\Big( \sqrt{x-\dfrac{1}{x} }-1\Big) ^2 = 0

\sqrt{x-\dfrac{1}{x} } = 1

x^2-x-1 = 0

x = \dfrac{1\pm\sqrt{1+4} }{2}

Cancel negative root since x>0

Therefore

x=\dfrac{1+\sqrt{5} }{2}

Steven Zheng posted 7 hours ago

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