Question

If $\frac{3+\sqrt{5}}{2\sqrt{5}+3}=a+b\sqrt{5}$, then the values of irrational numbers a and b are

• A. $\frac{1}{11},\frac{3}{11}$
• B. $7,\frac{8}{11}$
• C. $5,11$
• D. $\frac{2}{11},\frac{3}{11}$

Answer 1

The correct option is A $\frac{1}{11},\frac{3}{11}$

$\frac{3+\sqrt{5}}{2\sqrt{5}+3}$can be simplified by rationalizing the denominator $(2\sqrt{5}+3)$. We can rationalize the given denominator by multiplying the given number $(2\sqrt{5}+3)$with its rationalizing factor $(2\sqrt{5}-3)$.

Multiplying the numerator and denominator with $(2\sqrt{5}-3)$, we get

$\frac{(3+\sqrt{5})(2\sqrt{5}-3)}{(2\sqrt{5}+3)(2\sqrt{5}-3)}=\frac{6\sqrt{5}-9+10-3\sqrt{5}}{20-9}=\frac{1+3\sqrt{5}}{20-9}=\frac{1}{11}+\frac{3}{11}\sqrt{5}\text{Hence, a}=\frac{1}{11}\text{and b}=\frac{3}{11}$