Question

If \sqrt{x} +\dfrac{2}{\sqrt{x} } = x-\dfrac{4}{x}

find the value of x +\dfrac{4}{x}

Collected in the board: Radical equations

Steven Zheng posted 20 hours ago

Answer

Apply difference of squares formula on right hand side

\sqrt{x} +\dfrac{2}{\sqrt{x} } = x-\dfrac{4}{x}

\sqrt{x} +\dfrac{2}{\sqrt{x} } =(\sqrt{x} +\dfrac{2}{\sqrt{x} })(\sqrt{x} -\dfrac{2}{\sqrt{x} })

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

Since \sqrt{x} +\dfrac{2}{\sqrt{x} }\ne 0

the only option to hold the equation true is

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

then

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

Square both sides

x+\dfrac{4}{x}-4 = 1

x+\dfrac{4}{x}- = 5

Steven Zheng posted 20 hours ago

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