Question

Find the value of b:

$\frac{\sqrt{3}+\sqrt{2}}{2}\sqrt{3}+\sqrt{2}8+b\sqrt{6}=?$

Answer 1

Consider the given equation,

$\frac{(\sqrt{3}+\sqrt{2})\sqrt{3}}{2}+\sqrt{28}+b\sqrt{6}=0$

$b\sqrt{6}=\frac{(\sqrt{3}+\sqrt{2})\sqrt{3}}{2}+\sqrt{28}$

$b=\frac{(\sqrt{3}+\sqrt{2})\sqrt{3}}{2\sqrt{6}}+\frac{\sqrt{28}}{\sqrt{6}}$

$b=\frac{(\sqrt{3}+\sqrt{2})\sqrt{3}}{2\sqrt{6}}+\frac{\sqrt{28}}{\sqrt{6}}$

$b=\frac{(\sqrt{3}+\sqrt{2})}{2\sqrt{2}}+\frac{\sqrt{14}}{\sqrt{3}}$

Hence, this is the answer.