Question

If the value of the expression $\frac{x-12}{\sqrt{8}}$ can be approximated to be equal to $\frac{x\sqrt{2}}{4}-C$, then determine the value of $C$ in the equation.

• A. $-3$
• B. $-\sqrt{3}$
• C. $2$
• D. $3\sqrt{2}$

Answer 1

Answer: (D)

$3\sqrt{2}$

Separating denominators, $\frac{x+12}{\sqrt{8}}$ $=$ $\frac{x}{\sqrt{8}}+\frac{12}{\sqrt{8}}$

$=$ $\frac{x}{2\sqrt{2}}+\frac{12}{2\sqrt{2}}$

Multiply $\sqrt{2}$ to upper and lower, we get

$=$ $\frac{x\sqrt{2}}{2\times 2}+\frac{12\sqrt{2}}{2\times 2}$

$=$ $\frac{x\sqrt{2}}{4}+\frac{12\sqrt{2}}{4}$

$=$ $\frac{x\sqrt{2}}{4}+3\sqrt{2}$

But given $\frac{x\sqrt{2}}{4}+3\sqrt{2}\ge \frac{x\sqrt{2}}{4}+C$

Then seeing both side $C=3\sqrt{2}$