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

Question

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

Uniteasy Admin · Accepted 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
$$x-4\sqrt{x}  =4-4\cdot 2 =-4$$

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