Question

if x=2+root3/2-root3 show that x^2-14x+1=0

Answer 1

here is ur answer first find the value of x thereafter the roots comes out to be in rational form so we know that in a quadratic equation if one root is rational then other root is also rational then we find root by putting (x-a)(x+a)

check out

Answer 2

Hope u like my process.
=====================
$= > x = \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \\ \\ or. \: \: x = \frac{ {(2 + \sqrt{3} )}^{2} }{(2 + \sqrt{3})(2 - \sqrt{3}) } \\ \\ or. \: \: x = \frac{ {2}^{2} + {( \sqrt{3}) }^{2} + 2 \times 2 \times \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} } \\ \\ or. \: \: x = \frac{4 + 3 + 4 \sqrt{3} }{4 - 3} = 7 + 4 \sqrt{3}$
Now..

$= > {x}^{2} - 14x + 1 \\ \\ = {(7 + 4 \sqrt{3} )}^{2} - 14(7 + 4 \sqrt{3} ) + 1 \\ \\ = {7}^{2} + {(4 \sqrt{3}) }^{2} + 2 \times 7 \times 4 \sqrt{3} - 14 \times 7 - 14 \times 4 \sqrt{3} + 1 \\ \\ = 49 + 48 + 56 \sqrt{3} - 98 - 56 \sqrt{3 } + 1 \\ \\ = 98 - 98 = 0..... < proved >$
_-_-_-_-_-_-_-_-_-_-_-
Hope this is ur required answer

Proud to help you