Question

anyone help me this question please give correct answer
if root2-root3/3root2+2root3=a+broot6 where a and b are rational numbers, then find the values of a and b.​

Answer 1

Answer:

$\blue{a=12/6}$

$\green{b=-5/6}$

Step-by-step explanation:

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

Rationalizing the LHS

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

using the identity

$\green{(a+b)(a-b)=a^2-b^2}$

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

$\frac{12 - 5 \sqrt{6} }{18 - 12} = a + b \sqrt{6}$

$\frac{12 - 5 \sqrt{6} }{6} = a + b \sqrt{6}$

$\frac{12}{6} - \frac{5 \sqrt{6} }{6} = a + b \sqrt{6}$

$a = \frac{12}{6} \:$

$b = - \frac{5}{6}$

Answer 2

Answer:

Hey mate plzz refer to the attachment