Question

If a and b are rational numbers, find 'a' and 'b' when

$\frac{\sqrt{7}-2}{\sqrt{7}+2}=a\sqrt{7}+b$

• A. $a=-\frac{4}{3},b=\frac{-11}{3}$
• B. $a=\frac{-4}{3},b=\frac{11}{3}$
• C. $a=\frac{4}{3},b=\frac{-11}{3}$
• D. $a=\frac{4}{3},b=\frac{11}{3}$

Answer 1

The correct option is B

$a=\frac{-4}{3},b=\frac{11}{3}$

$\frac{\sqrt{7}-2}{\sqrt{7}+2}=a\sqrt{7}+b$

$\frac{\sqrt{7}-2}{\sqrt{7}+2}\times \frac{\sqrt{7}-2}{\sqrt{7}-2}=a\sqrt{7}+b$

$\frac{(\sqrt{7}-2{)}^2}{(\sqrt{7}{)}^2-(2{)}^2}=a\sqrt{7}+b$

The numerator is of the form $(a-b{)}^2$ and the denominator is of the form $a^2-b^2$

$\therefore \frac{(\sqrt{7}{)}^2-2(\sqrt{7}\left)\right(2)+(2{)}^2}{7-4}=a\sqrt{7}+b$

$\frac{7-4\sqrt{7}+4}{3}=a\sqrt{7}+b$

$-\frac{4}{3}\sqrt{7}+\frac{11}{3}=a\sqrt{7}+b$

$\text{Comparing we get,}a=-\frac{4}{3},b=\frac{11}{3}$