Question

find the zeros of the following quadratic polynomials and verify the relationship between the zeroes and coefficients 6x²-3-7x​

Answer 1

Step-by-step explanation:

$6 {x}^{2} - 7x - 3 = 0 \\ x = \frac{18}{12} \: \: or \: \: x = \frac{ - 4}{12} \\ x( \alpha ) = \frac{3}{2} \: \: or \: \: x ( \beta )= \frac{ - 1}{3} \\ \\ \alpha + \beta = \frac{ - b}{a} \\ \frac{3}{2} - \frac{1}{3} = \frac{7}{6} \\ \\ \alpha \beta = \frac{c}{a} \\ \frac{3}{2} \times \frac{ - 1}{3} = \frac{ - 1}{2}$