Question

$\frac{3+\sqrt{12}-\sqrt{27}}{7+\sqrt{192}-\sqrt{48}}=a+b\sqrt{3}$, then find $a$ and $b$.

• A. $a=33,b=-19$
• B. $a=30,b=-9$
• C. $a=33,b=19$
• D. $a=33,b=9$

Answer 1

The correct option is A $a=33,b=-19$

$\frac{3+\sqrt{12}-\sqrt{27}}{7+\sqrt{192}-\sqrt{48}}=\frac{3+2\sqrt{3}-3\sqrt{3}}{7+8\sqrt{3}-4\sqrt{3}}=\frac{3-\sqrt{3}}{7+4\sqrt{3}}=\frac{3-\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}=\frac{33-19\sqrt{3}}{7^2-{\left(4\sqrt{3}\right)}^2}=33-19\sqrt{3}$

On comparing we get $a=33$ and $b=-19$

SO corrct answer is option A