Question

find the values of a and b if 3/(3+2root2) + 2/(3-2root2) = a+b root 2....explain​

Answer 1

$\large\underline{\sf{Given- }}$

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

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:a \: and \: b$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{3}{3 + 2 \sqrt{2} } + \dfrac{2}{3 - 2 \sqrt{2} } = a + b \sqrt{2} - - (1)$

Consider,

$\blue{\bf :\longmapsto\:\dfrac{3}{3 + 2 \sqrt{2} }}$

On rationalizing the denominator, we get

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

$\rm \: = \: \: \dfrac{9 - 6 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} }$

$\rm \: = \: \: \dfrac{9 - 6 \sqrt{2} }{9 - 8}$

$\rm \: = \: \: \dfrac{9 - 6 \sqrt{2} }{1}$

$\rm \: = \: \: 9 - 6 \sqrt{2}$

$\blue{\bf\implies \:\dfrac{3}{3 + 2 \sqrt{2} } = 9 - 6 \sqrt{2}} - - - (2)$

Consider,

$\red{\bf :\longmapsto\:\dfrac{2}{3 - 2 \sqrt{2} }}$

On rationalizing the denominator, we get

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

$\rm \: = \: \: \dfrac{6 + 4 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2} )}^{2} }$

$\rm \: = \: \: \dfrac{6 + 4 \sqrt{2} }{9 - 8}$

$\rm \: = \: \: \dfrac{6 + 4\sqrt{2} }{1}$

$\rm \: = \: \: 6 + 4 \sqrt{2}$

$\red{\bf :\longmapsto\:\dfrac{2}{3 - 2 \sqrt{2} } = 6 + 4 \sqrt{2} } - - - (3)$

Consider,

$\bf :\longmapsto\:\dfrac{3}{3 + 2 \sqrt{2} } + \dfrac{2}{3 - 2 \sqrt{2} } = a + b \sqrt{2}$

On substituting the value from equation (2) and (3), we get

$\rm :\longmapsto\:9 - 6 \sqrt{2} + 6 + 4 \sqrt{2} = a + b \sqrt{2}$

$\bf :\longmapsto\: \red{15} \: - \: \blue{5 \sqrt{2}} \: = \: \red{a} \: + \: \blue{ b \sqrt{2} }$

On comparing, we get

$\bf :\longmapsto\:a = 15 \: \: \: \: and \: \: \: \: \: b = - \: 5$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

Answer:

a = 15

b = -2

Step-by-step explanation:

Given:

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

To find:

The values of a and b.

Solution:

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

$\implies \sf{\dfrac{3(3 - 2\sqrt{2}) + 2(3 + 2\sqrt{2})}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})}} = a+b\sqrt{2}}$

$\implies \sf{\dfrac{(9 - 6\sqrt{2}) + (6 + 4\sqrt{2})}{(3)^2 - (2\sqrt{2})^2} = a+b\sqrt{2}}$

$\implies \sf{\dfrac{9 - 6\sqrt{2} + 6 + 4\sqrt{2}}{9 - 8} = a+b\sqrt{2}}$

$\implies \sf{\dfrac{15 - 2\sqrt{2}}{1} = a+b\sqrt{2}}$

$\implies \sf{15 - 2\sqrt{2} = a+b\sqrt{2}}$

Comparing LHS and RHS, we get

a = 15

b = -2