Question

√2+√3/3√2-2√3
= a-bro, find value of a, b.​

Answer 1

Correct question :

√2+√3/3√2-2√3  = a-b√6

Answer :

a = 2, b = -5/6

Step-by-step explanation :

   $\underline{\underline{\bf Rationalizing \ factor :}}$

   $\bigstar$  The factor of multiplication by which rationalization is done, is called as rationalizing factor.

   $\bigstar$  If the product of two surds is a rational number, then each surd is a rationalizing factor to other.

   $\bigstar$  For example, rationalizing factor of (3 + √2) is (3 - √2)

_____________________________

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

Rationalizing factor = 3√2 + 2√3

$\bf =\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}} \times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} \\\\\\ =\frac{(\sqrt{2}+\sqrt{3})(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})} \\\\\\=\frac{\sqrt{2}(3\sqrt{2}+2\sqrt{3})+\sqrt{3}(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})} \\\\\\=\frac{3(\sqrt{2})^2+2\sqrt{6}+3\sqrt{6}+2(\sqrt{3})^2}{(3\sqrt{2})^2-(2\sqrt{3})^2} \\\\\\=\frac{3(2)+5\sqrt{6}+2(3)}{9(2)-4(3)} \\\\\\=\frac{6+5\sqrt{6}+6}{18-12}$

$\bf =\frac{12+5\sqrt{6}}{6} \\\\\\ =\frac{12}{6} +\frac{5\sqrt{6} }{6} \\\\\\ =2+\frac{5}{6} \sqrt{6}\\\\\\ =2-(\frac{-5}{6}\sqrt{6})\\\\\\ =a-b\sqrt{6}$

∴ a = 2, b = -5/6