Question

Please solve ques no. 15.
From Chapter-2 Polynomial

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Answer 1

Hello,


The form of a quadratic equation is

$k( {x}^{2} - ( \alpha + \beta )x + \alpha \beta )$
So,

Given that
$\alpha = \frac{3 - \sqrt{3} }{5} \\ \\ \beta = \frac{3 + \sqrt{3} }{5}$


Finding the sum and the product of the quadratic equation's zeroes.

$\alpha + \beta = \frac{3 - \sqrt{3} }{5} + \frac{3 + \sqrt{3} }{5} \\ \\ = \frac{3 - \sqrt{3} + 3 + \sqrt{3} }{5} \\ \\ = \frac{6}{5}$
$\alpha \beta = ( \frac{3 - \sqrt{3} }{5} )( \frac{3 + \sqrt{3} }{5} ) \\ \\ = \frac{(3 - \sqrt{3})(3 + \sqrt{3} )}{25} \\ \\ = \frac{ {3}^{2} - {( \sqrt{3}) }^{2} }{25} \\ \\ = \frac{9 - 3}{25} \\ \\ = \frac{6}{25}$
Now

Putting the values of the following in the equation form

$k( {x}^{2} - \frac{6}{5} x + \frac{6}{25} ) \\ \\ = (25 {x}^{2} - 30x + 6)$
When the value of the k is equal to 25.


Hope this will be helping you.....