Question

what would be the answer

Answer 1

Hope u like my process
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Rationalizing terms to get a short value of x.

$= > \: \: x = \frac{ \sqrt{a + 2b} + \sqrt{a - 2b} }{ \sqrt{a + 2b} - \sqrt{a - 2b} } \\ \\ or. \: \: x = \frac{ {( \sqrt{a + 2b} + \sqrt{a - 2b} )}^{2} }{( \sqrt{a + 2b} - \sqrt{a + 2b})( \sqrt{a + 2b} + \sqrt{a - 2b} ) } \\ \\ or. \: \: \: x = \frac{((a + 2b) +(a - 2b) + 2 \sqrt{(a + 2b)(a - 2b)} )}{ {( \sqrt{a + 2b}) }^{2} - {( \sqrt{a - 2b}) }^{2} } \\ \\ or. \: \: \: x = \frac{a + 2b + a - 2b + 2 \sqrt{ {a}^{2} - 4 {b}^{2} } }{a + 2b - a + 2b} \\ \\ or. \: \: \: x = \frac{2a + 2 \sqrt{ {a}^{2} - 4 {b}^{2} } }{4b} \\ \\ or. \: \: x = \frac{a + \sqrt{ {a}^{2} - 4 {b}^{2} } }{2b} ......(1)$
Now.
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By Shreedhar Acharya's formula
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For equation like,

=> mx² - nx + o = 0 ___(3)

$x = \frac{ + n - \sqrt{ {n}^{2} - 4mo } }{2m} ........(2)$

So,
comparing our's equation (1) to the following Shreedhar Acharya's equation.(2).

We get,

=> m = b ; n = a ; o = b

So, putting the values in equation (3),we get,

=> bx² - ax + b = 0___(proved).

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