Question

ANSWER THIS QUESTION TO GET THE BRAINLIEST.

Answer 1

Given :-

$\dfrac{3 + \sqrt{8} }{3 - \sqrt{8} } + \dfrac{3 - \sqrt{8} }{3 + \sqrt{8} } = a + b \sqrt{8}$

To find :-

• Value of a, b

SOLUTION:-

Take L.C.M to the denominators

$\dfrac{(3 + \sqrt{8})(3 + \sqrt{8}) + (3 - \sqrt{8})(3 - \sqrt{8} ) }{(3 - \sqrt{8})(3 + \sqrt{8}) }$

$\dfrac{(3 + \sqrt{8}) {}^{2} + (3 - \sqrt{8} ) {}^{2} }{(3 + \sqrt{8} )(3 - \sqrt{8}) }$

Numerator is in form of (a+b)² +(a-b)² = 2a² + 2b²

Denominator in form of (a+b)(a-b) = a²-b²

By applying these formulae we get,

$\dfrac{2(3) {}^{2} + 2( \sqrt{8}) {}^{2} }{(3) {}^{2} - ( \sqrt{8}) {}^{2} }$

$\dfrac{2(9) + 2(8)}{9 - 8}$

$18 + 16$

$34$

$34 = a + b \sqrt{8}$

$17 + 17 = a + b \sqrt{8}$

Multiply and divide by 8

$(\dfrac{17}{8} + \dfrac{17}{8} ) \sqrt{8} \times \sqrt{8} = a + b \sqrt{8}$

$\dfrac{17 \sqrt{8} }{8} + \dfrac{17 \sqrt{8} }{8} \times \sqrt{8} = a + b \sqrt{8}$

$\dfrac{17}{ \sqrt{8} } + \dfrac{17}{ \sqrt{8} } \times \sqrt{8} = a + b \sqrt{8}$

So,

$a = \dfrac{17}{ \sqrt{8} } \: b = \dfrac{17}{ \sqrt{8} }$

Answer 2

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