Question

Find the conjugate of $\sqrt{\frac{4}{3}}-\sqrt{\frac{3}{4}}$ and simplify it.

• A. $\frac{7}{\sqrt{12}}$
• B. $-\frac{7}{\sqrt{12}}$

Answer 1

Answer: (A), (B)

$-\frac{7}{\sqrt{12}}$

Conjugate of $\sqrt{\frac{4}{3}}-\sqrt{\frac{3}{4}}\text{is}\sqrt{\frac{4}{3}}+\sqrt{\frac{3}{4}}.$

$\sqrt{\frac{4}{3}}+\sqrt{\frac{3}{4}}=\frac{\sqrt{4}\times \sqrt{4}+\sqrt{3}\times \sqrt{3}}{\sqrt{3}\times \sqrt{4}}=\frac{4+3}{\sqrt{3\times 4}}=\frac{7}{\sqrt{12}}$

Also,
$\sqrt{\frac{4}{3}}-\sqrt{\frac{3}{4}}=-\sqrt{\frac{3}{4}}+\sqrt{\frac{4}{3}}$

$\text{Conjugate of}-\sqrt{\frac{3}{4}}+\sqrt{\frac{4}{3}}\text{is}-\sqrt{\frac{3}{4}}-\sqrt{\frac{4}{3}}.$

After simplification, $-\sqrt{\frac{3}{4}}-\sqrt{\frac{4}{3}}$ results to $-\frac{7}{\sqrt{12}}$