Question

${x }^{2} - x \times \sqrt{3} + 4$
FIND VALUE OF X

Answer 1

In order to find the value of x , we have to find the zeros of the given polynomial.

We shall solve it by completing square method ,

$p(x) = {x}^{2} - \sqrt{3} x + 4 \\ \\ \implies {x}^{2} - \sqrt{3} x + 4 = 0 \\ \\ \implies {x}^{2} - \sqrt{3} x = - 4 \\ \\ \implies {x}^{2} - 2 \times \frac{ \sqrt{3} }{2} \times x = - 4 \\ \\ \implies {x}^{2} - 2 \times \frac{ \sqrt{3} }{2} \times x + { (\frac{ \sqrt{3} }{2} )}^{2} = - 4 \\ \\ \implies {(x - \frac{ \sqrt{3} }{2} )}^{2} = - 4 \\ \\ \implies x - \frac{ \sqrt{3} }{2} = \pm2 \\ \\ \implies x = \frac{ \sqrt{3} }{2} + 2 \: or \: \frac{ \sqrt{3} }{2} - 2 \\ \\ \implies x = \frac{ \sqrt{3} + 4}{2} \: or \: \frac{ \sqrt{3} - 4 }{2}$

Or Else we can solve it by using formula of discriminant.

D = b^2 - 4ac

x = ( b ± rootD)/ 2a