Question

Expand :
${[\sqrt{3x}-\frac{\sqrt{2}}{5}y]}^2$

• A. $3x^2-\frac{2}{5}\sqrt{6x}+\frac{2}{25}$
• B. $3x-\frac{2\sqrt{6x}y}{5}+\frac{2}{25}{y}^2$
• C. $3x^2-\sqrt{6x}+\frac{4}{25}$
• D. $x^2-\frac{2}{5}\sqrt{6x}+\frac{4}{25}$

Answer 1

The correct option is B $3x-\frac{2\sqrt{6x}y}{5}+\frac{2}{25}{y}^2$

$\text{Given :}{[\sqrt{3x}-\frac{\sqrt{2}}{5}y]}^2$

Using the identity,
$(a-b{)}^2=a^2-2ab+b^2$,

$\text{we have,}a=\sqrt{3x}\text{and}b=\frac{\sqrt{2}}{5}y.$

So,
${[\sqrt{3x}-\frac{\sqrt{2}}{5}y]}^2$
$=(\sqrt{3x}{)}^2-2(\sqrt{3x}\times \frac{\sqrt{2}}{5}y)+(\frac{\sqrt{2}}{5}y{)}^2$
$=3x-\frac{2\sqrt{6x}y}{5}+\frac{2}{25}{y}^2$