Question

If 'p' and 'q' are rational numbers, then find 'p' and 'q' given that $\frac{3+\sqrt{7}}{3-\sqrt{7}}=p+q\sqrt{7}$

• A. $p=-8,q=3$
• B. $p=-8,q=-3$
• C. $p=8,q=-3$
• D. $p=8,q=3$

Answer 1

The correct option is D

$p=8,q=3$

$\frac{3+\sqrt{7}}{3-\sqrt{7}}=p+q\sqrt{7}$

$\frac{3+\sqrt{7}}{3-\sqrt{7}}\times \frac{3+\sqrt{7}}{3-\sqrt{7}}=p+q\sqrt{7}$

$\frac{(3+\sqrt{7}{)}^2}{(3{)}^2-(\sqrt{7}{)}^2}=p+q\sqrt{7}$

$\frac{(3{)}^2+2(3\left)\right(\sqrt{7})+(\sqrt{7}{)}^2}{9-7}=p+q\sqrt{7}$

$\frac{9+6\sqrt{7}+7}{2}=p+q\sqrt{7}$

$\frac{16+6\sqrt{7}}{2}=p+q\sqrt{7}$

$\frac{\text{/}2(8+3\sqrt{7})}{\text{/}2}=p+q\sqrt{7}$

$p=8,q=3$