Question

If $\frac{4}{2+\sqrt{3}+\sqrt{7}}=\sqrt{a}+\sqrt{b}-\sqrt{c}$, then which of the following can be true-

• A. $a=\frac{4}{3},b=1,c=\frac{7}{3}.$
• B. $a=1,b=2,c=0$
• C. $a=\frac{1}{3},b=0,c=\frac{7}{3}.$
• D. $a=4,b=1,c=\frac{1}{3}.$

Answer 1

The correct option is A $a=\frac{4}{3},b=1,c=\frac{7}{3}.$

$\frac{4}{2+\sqrt{3}+\sqrt{7}}=\sqrt{a}+\sqrt{b}-\sqrt{c}$

$LHS=\frac{4}{2+\sqrt{3}+\sqrt{7}}\times \frac{2+\sqrt{3}-\sqrt{7}}{2+\sqrt{3}-\sqrt{7}}$

$=\frac{4(2+\sqrt{3}-\sqrt{7})}{4+3+4\sqrt{3}-7}$

$=\frac{4(2+\sqrt{3}-\sqrt{7})}{4\sqrt{3}}$

$=\sqrt{\frac{4}{3}}+\sqrt{1}-\sqrt{\frac{7}{3}}$

$OncomparingwithRHS,weget$

$a=\frac{4}{3},b=1,c=\frac{7}{3}.$