Question

my question plz solve it

Answer 1

$4 \sqrt{3} {x}^{2} + 5x - 2 \sqrt{3} \\ 4 \sqrt{3} {x}^{2} + 8x - 3x - 2 \sqrt{3} \\ 4x( \sqrt{3}x + 2) - \sqrt{3} ( \sqrt{3} x + 2) \\ ( \sqrt{3}x + 2)(4x - \sqrt{3} )$
to find zeros equate the factors with zero
$\sqrt{3} x + 2 = 0 \\ x = \frac{ - 2}{ \sqrt{3} } \\ 4x - \sqrt{3} = 0 \\ x = \frac{ \sqrt{3} }{4}$
from coefficient of polynomial,sum of zeros and product of zeros
$\alpha + \beta = - \frac{b}{a} \\ = \frac{ - 5}{4 \sqrt{3} } \\ \alpha \beta = \frac{c}{a} = \frac{-2 \sqrt{3} }{4 \sqrt{3} } = \frac{-1}{2}$
from zeros: sum and product of zeros
$= \frac{ - 2}{ \sqrt{3} } + \frac{ \sqrt{3} }{4 } \\ = \frac{ - 8 + 3}{4 \sqrt{3} } \\ = \frac{ - 5}{4 \sqrt{3} } \\ \alpha \beta = ( \frac{ - 2}{ \sqrt{3} } ) \times ( \frac{ \sqrt{3} }{4} ) = - \frac{2 \sqrt{3} }{4 \sqrt{3} } \\ = \frac{ - 1}{2}$