Question

3+ 2 root 2 / 3 minus root 2 is equal to a + b root 2 where A and B are rational number find the values of a and b​

Answer 1

Given:

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

To find:

The values of a and b

Solution:

Rationalising the denominator

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

We know that

(a-b)(a+b) = (a²-b²)

and

(a+b)² = a² +b² +2ab

$= \frac{(3) {}^{2} + ( \sqrt{2} ) {}^{2} + 2 \times 3 \times ( \sqrt{2}) }{9 - 2} \\ \\ = \frac{9 + 2 + 6 \sqrt{2} }{7} \\ \\ = \frac{11 + 6 \sqrt{2} }{7}$

Also, we have

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2} \\ \\ \ \: Therefore\\ \\ \large \implies \frac{11 + 6 \sqrt{2} }{7} = a + b \sqrt{2}$

Now comparing them

$\implies \: \frac{11}{7} + \frac{6 }{7} \sqrt{2} = a + b \sqrt{2} \\ \\ On \: comparing \\ \\ \implies \: a = \frac{11}{7} \\ \\ \implies \: b = \frac{6}{7}$

Hence, the value of a = 11/7 and b = 6/7.

Answer 2

Answer:

conjucate of this is 3+rooot2/3+rooot 2

3+root2 the whole square/3_root 2 3+root 2

as per formula

a+b the whole square/a square _bsquare formula

we get the answer as 11+6 root 2

Step-by-step explanation:

1. hope it helps u❤