Question

if √3+√2/√3-√2 = a+b√6 then find the values of a and b​

Answer 1

Answer:

$\mathrm{solve\:for\:a,\:\sqrt{3}+\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{2}=a+b\sqrt{6}\quad :\quad a=\sqrt{3}+\sqrt{\frac{2}{3}}-\sqrt{2}-b\sqrt{6}}$

Step-by-step explanation:

$\sqrt{3}+\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{2}=a+b\sqrt{6}$

$a+b\sqrt{6}=\sqrt{3}+\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{2}$

$\mathrm{Combine\:same\:powers}\::\quad \frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$

$a+b\sqrt{6}=\sqrt{3}+\sqrt{\frac{2}{3}}-\sqrt{2}$

$\mathrm{Subtract\:}b\sqrt{6}\mathrm{\:from\:both\:sides}$

$a+b\sqrt{6}-b\sqrt{6}=\sqrt{3}+\sqrt{\frac{2}{3}}-\sqrt{2}-b\sqrt{6}$

$a=\sqrt{3}+\sqrt{\frac{2}{3}}-\sqrt{2}-b\sqrt{6}$

Answer 2

Step-by-step explanation:

$= > \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

Multiply and divide with "√3 + √2"

$= > \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: \times \: \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$=> \frac{(√3+√2)²}{(√3)²~-~(√2)²}$

$=> \frac{(√3)² + (√2)² + 2√3√2)}{(3~-~2)}$

$=> \frac{(3~+~2~+2√6)}{1}$

$=> \frac{5~+2√6}{1}$

:. a + b√6 = 5 + 2√6

a = 5

b = 2