Question

if a and b are rational numbers than find.
their value in
=√3-√2/√3+√2=a+b√6
​

Answer 1

Given:

The equation -

• $\bf \longmapsto \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} = a+b\sqrt{6}$

What To Find:

We have to find -

• The value of a and b.

Solution:

$\sf \longmapsto \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} = a+b\sqrt{6}$

Rationalise the denominator of LHS by its conjugate.

$\sf \longmapsto \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}= a+b\sqrt{6}$

Take them as common,

$\sf \longmapsto \dfrac{(\sqrt{3}-\sqrt{2}) \times (\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2}) \times (\sqrt{3}-\sqrt{2})} = a+b\sqrt{6}$

Solve the numerator as,

$\sf \longmapsto \dfrac{\sqrt{3}(\sqrt{3}-\sqrt{2}) -\sqrt{2} (\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2}) \times (\sqrt{3}-\sqrt{2})} = a+b\sqrt{6}$

Solve the first brackets,

$\sf \longmapsto \dfrac{3-\sqrt{6} -\sqrt{2} (\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2}) \times (\sqrt{3}-\sqrt{2})} = a+b\sqrt{6}$

Solve the second brackets,

$\sf \longmapsto \dfrac{3-\sqrt{6} -\sqrt{6}+2}{(\sqrt{3}+\sqrt{2}) \times (\sqrt{3}-\sqrt{2})} = a+b\sqrt{6}$

Rearrange the terms,

$\sf \longmapsto \dfrac{3+2-\sqrt{6} -\sqrt{6}}{(\sqrt{3}+\sqrt{2}) \times (\sqrt{3}-\sqrt{2})} = a+b\sqrt{6}$

Solve the terms,

$\sf \longmapsto \dfrac{5-2\sqrt{6}}{(\sqrt{3}+\sqrt{2}) \times (\sqrt{3}-\sqrt{2})} = a+b\sqrt{6}$

Solve the denominator using the identity (a + b) (a - b) = a² - b²

$\sf \longmapsto \dfrac{5-2\sqrt{6}}{(\sqrt{3})^2-(\sqrt{2})^2} = a+b\sqrt{6}$

Find the square,

$\sf \longmapsto \dfrac{5-2\sqrt{6}}{3-2} = a+b\sqrt{6}$

Subtract the values,

$\sf \longmapsto \dfrac{5-2\sqrt{6}}{1} = a+b\sqrt{6}$

Can be written as,

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

By inspection, we can see that -

• $\sf \longmapsto 5 = a$

• $\sf \longmapsto -2 = b$

Final Answer:

∴ Thus, the value of a and b are 5 and - 2 respectively where 5 and - 2 are rational numbers.