If √{{19 + 8√3}/{7 - 4√3}} = a+b√3, then a equals

Question

If \(\sqrt{\cfrac{19+8\sqrt3} {7-4\sqrt3}} =a+b\sqrt3,\) then a equals

(a) 6

(b) 4

(c) 11

(d) 7

Answer 1

Correct option : (c) 11

\(\because\) \(19+8\sqrt3 =16+3+8\sqrt3\)

\(=4^2 +(\sqrt3)^2 +2 \times 4\times \sqrt3\)

\(=(4+\sqrt3)^2\)

Also \(7-4\sqrt3 =4+3-4\sqrt3\)

\(=2^2+(\sqrt3)^2 -2 \times 2\times\sqrt3\)

\(=(2-\sqrt3)^2\)

\(\therefore\) \(\sqrt{\frac{19+8\sqrt3}{7-4\sqrt3}} =\frac{4+\sqrt3}{2-\sqrt3} =\frac{4+\sqrt3}{2-\sqrt3} \times \frac{2+\sqrt3}{2+\sqrt3}\)

\(=\frac{8+2\sqrt3 +4\sqrt3+3}{4-(\sqrt3)^2}\)

\( =\frac{11+6\sqrt3}{1} =11+6\sqrt3\)

\(=a+b\sqrt3\)

\(\therefore \boxed{a=11},b=6\).