Question

In the following question find a and b

Answer 1

Step-by-step explanation:

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Answer 2

✓Question :

$\frac{3 + \sqrt{8} }{3 - \sqrt{8} } + \frac{3 - \sqrt{8} }{3 + \sqrt{8} } = a + b \sqrt{2}$

✓ To find :

» values of a and b

✓ Solution

$\frac{3 + \sqrt{8} }{3 - \sqrt{8} } + \frac{3 - \sqrt{8} }{3 + \sqrt{8} }$

$= \frac{(3 + \sqrt{8 })(3 + \sqrt{8}) \: \: \: \: + (3 - \sqrt{8})(3 - \sqrt{8} ) } {(3 - \sqrt{8})(3 + \sqrt{8}) }$

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

$\frac{ = 9 + 8 + 6 \sqrt{8} + 9 + 8 - 6 \sqrt{8} }{9 - 8}$

$= \frac{17 + 17}{1} = 34$

$\therefore \: \frac{3 + \sqrt{8} }{3 - \sqrt{8} } + \frac{3 - \sqrt{8} }{3 + \sqrt{8} } = 34$

$a + b \sqrt{2} = 34 + 0$

$by \: comparing \: a = 34 \: and \: b = 0$

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More info :

✓ Here we need to take LCM [ Least common Multiple ] first and then solve the given condition. After that we have to equate it with the RHS to find the values of unknown variable 'a' and 'b'.