Question

if √3+√2/√3-√2=a+b√6 then show that a=5,b=1.​

Answer 1

Answer:

{a, b} = {5 , 2}

Step-by-step explanation:

$\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}=a+b\sqrt6$

$\to \frac{(\sqrt3+\sqrt2)(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}=a+b\sqrt6$

$\to \frac{(\sqrt3+\sqrt2)^{2}}{((\sqrt3)^2-(\sqrt2)^2}=a+b\sqrt6$

$\to \frac{(\sqrt3)^2+(\sqrt2)^2+2\sqrt3\sqrt2}{(3-2)}=a+b\sqrt6$

$\to \frac{3+2+2\sqrt{3\times2}}{1}=a+b\sqrt6$

$\to 5+2\sqrt{6}=a+b\sqrt6$

so, a = 5 and b = 2   (comparing rational and irrational part)