Question

find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficient


√3x2 - 8x + 4√3​

Answer 1

Hey

$f(x) = \sqrt{3} x {}^{2} - 8x + 4 \sqrt{ 3} = \sqrt{3} x {}^{2} - 6x - 2x + 4 \sqrt{3} = (x - 2 \sqrt{3} )( \sqrt{3} x - 2) \\ \\ f(x) = 0 \\ = > (x - 2 \sqrt{3} )( \sqrt{3} x - 2) = 0 \\ = > x = 2 \sqrt{3} \: \: or \: \: \frac{2}{ \sqrt{3} } \\ \\ = > \alpha = 2 \sqrt{3} \: \: and \: \: \beta = \frac{2}{ \sqrt{3} } \\ \\ Sum \: of \: zeros :\\ \alpha + \beta = 2 \sqrt{3} + \frac{2}{ \sqrt{3} } = \frac{6 + 2}{ \sqrt{3} } = \frac{8}{ \sqrt{3} } = - \frac{b}{a} \\ \\ Product \: of \: zeros :\\ \alpha \beta = 2 \sqrt{3} \times \frac{2}{ \sqrt{3} } = 4 = \frac{c}{a}$