Question

√3+√2/√3-√2=a+b√6.find the value of a, b

Answer 1

Answer:

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

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

Rationalising factor = $\sqrt{3} + \sqrt{2}$

$\frac{\sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2} }$   X $\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}$  =   $\frac{\sqrt{3}+ \sqrt{2} (\sqrt{3}+ \sqrt{2}) }{\sqrt{3}- \sqrt{2}(\sqrt{3}+ \sqrt{2})}$

                            = $\frac{(\sqrt{3}+ \sqrt{2})^{2} }{\sqrt{3}^{2} - (\sqrt{2} )^{2} }$

                            = $\frac{(\sqrt{3})^{2}+ 2 X \sqrt{3} +(\sqrt{2} )^{2} }{3 -2 }$

                            = $\frac{3 + 2\sqrt{6}+2 }{1}$

                            = $5 + 2\sqrt{6}$

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

$5 + 2\sqrt{6}$  $= a +b\sqrt{6}$

a = 5, b = 2

∴ a = 5, b = 2.