Question

find the value of a and b such that 3+√2÷3-√2=a+b√2​

Answer 1

Solution:-

Given:-

$\rm \implies \: \dfrac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2}$

To Find the value of a and b

Now take

$\rm \implies \: \dfrac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \dfrac{3 + \sqrt{2} }{3 + \sqrt{2} }$

$\rm \implies \: \dfrac{(3 + \sqrt{2} )(3 + \sqrt{2} )}{(3 - \sqrt{2} )(3 + \sqrt{2}) }$

Using this identities

$\rm \implies \: (a - b)(a + b) = {a}^{2} - {b}^{2}$

$\rm \implies \: (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab$

We get

$\rm \implies \: \dfrac{(3 + \sqrt{2} ) {}^{2} }{(3) {}^{2} - ( \sqrt{2}) {}^{2} }$

$\rm \implies \dfrac{9 + 2 + 2 \times 3 \times \sqrt{2} }{9 - 2}$

$\implies \: \dfrac{11 + 6 \sqrt{2} }{7}$

$\implies \: \dfrac{11}{7} + \dfrac{6}{7} \sqrt{2}$

Now compare with

$\rm \implies \: a + b \sqrt{2}$

We get

$\implies \rm \: a = \dfrac{11}{7} \: and \: b = \dfrac{6}{7}$

Answer 2

Given:-

\rm \implies \: \dfrac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2}⟹

3−

2

3+

2

=a+b

2

To Find the value of a and b

Now take

\rm \implies \: \dfrac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \dfrac{3 + \sqrt{2} }{3 + \sqrt{2} }⟹

3−

2

3+

2

×

3+

2

3+

2

\rm \implies \: \dfrac{(3 + \sqrt{2} )(3 + \sqrt{2} )}{(3 - \sqrt{2} )(3 + \sqrt{2}) }⟹

(3−

2

)(3+

2

)

(3+

2

)(3+

2

)

Using this identities

\rm \implies \: (a - b)(a + b) = {a}^{2} - {b}^{2}⟹(a−b)(a+b)=a

2

−b

2

\rm \implies \: (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab⟹(a+b)

2

=a

2

+b

2

+2ab

We get

\rm \implies \: \dfrac{(3 + \sqrt{2} ) {}^{2} }{(3) {}^{2} - ( \sqrt{2}) {}^{2} }⟹

(3)

2

−(

2

)

2

(3+

2

)

2

\rm \implies \dfrac{9 + 2 + 2 \times 3 \times \sqrt{2} }{9 - 2}⟹

9−2

9+2+2×3×

2

\implies \: \dfrac{11 + 6 \sqrt{2} }{7}⟹

7

11+6

2

\implies \: \dfrac{11}{7} + \dfrac{6}{7} \sqrt{2}⟹

7

11

+

7

6

2

Now compare with

\rm \implies \: a + b \sqrt{2}⟹a+b

2

We get

\implies \rm \: a = \dfrac{11}{7} \: and \: b = \dfrac{6}{7}⟹a=

7

11

andb=

7

6