Question

If x+\dfrac{16}{\sqrt{x} } =12, find the value of x-4\sqrt{x}

Collected in the board: Radical equations

Steven Zheng posted 2 days ago

Answer

x+\dfrac{16}{\sqrt{x}} =16-4

x-16 = -\dfrac{16}{\sqrt{x}}-4

Using difference of squares formula

( \sqrt{x}-4 )(\sqrt{x}+4 ) = -\dfrac{4(4+\sqrt{x} )}{\sqrt{x} }

Cancel \sqrt{x}+4 since it is larger than 1

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

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

\sqrt{x} = 2

x=4

Finally

x-4\sqrt{x} =4-4\cdot 2 =-4

Steven Zheng posted 2 days ago

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