Question

if ✓3+✓2/✓3-✓2=a+b✓6, find the value of a and b​

Answer 1

Answer:

$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6} \\ \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{\sqrt{3} + \sqrt{2} }{\sqrt{3} + \sqrt{2} } = a + b \sqrt{6} \\ \frac{( \sqrt{3 } + \sqrt{2})^{2}}{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} } = a + b \sqrt{6} \\ \frac{( \sqrt{3})^{2} + {( \sqrt{2} )}^{2} + 2 \sqrt{3} \sqrt{2} }{3 - 2} = a + b \sqrt{6} \\ \frac{3 + 2 + 2 \sqrt{6} }{1} = a + b \sqrt{6} \\ 5 + 2 \sqrt{6} = a + b \sqrt{6} \\ \implies \: a = 5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: b = 2$

Answer 2

Answer:

Value of a is 5 and value of b is 2

Step-by-step explanation:

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

Consider LHS:

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

Applying componendo and dividendo theorem, we get

$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\\\\\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}\\\\ \frac{(\,\sqrt{3} + \sqrt{2})\,(\,\sqrt{3} + \sqrt{2})\,}{(\,\sqrt{3} - \sqrt{2})\,(\,\sqrt{3} + \sqrt{2})\,}\\\\ \frac{(\,\sqrt{3} + \sqrt{2})\,^2}{(\,\sqrt{3})\,^2 - (\,\sqrt{2})\,^2}\\\\\frac{(\,\sqrt{3})\,^2 + (\,\sqrt{2})\,^2 + 2(\,\sqrt{3})\,(\,\sqrt{2})\,}{3 - 2}\\\\ \frac{3 + 2 + 2\sqrt{6}}{1}\\\\\ 5 + 2\sqrt{6}$

Now, we have

LHS  = RHS

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

On comparing, we get

a = 5

b = 2