Question

find the zeroes of the quadratic polynomial 6x2-3-7x and verify the relationship between the zeroes and co-efficients​

Answer 1

Answer:

$6 {x}^{2} - 7x - 3 = 0 \: \: has \: zeros \: \alpha \: \: and \: \: \beta .$

$\alpha + \beta = \frac{7}{6} \: \: and \: \: \alpha \beta = - \frac{3}{6} = - \frac{1}{2}$

now,

$6 {x}^{2} - 9x + 2x - 3 = 0$

$3x(2x - 3) + 1(2x - 3) = 0$

$(2x - 3)(3x + 1) = 0$

$x = \frac{3}{2} \: \: or \: \: - \frac{1}{3}$

$\frac{3}{2} + ( - \frac{1}{3} ) = \frac{9 - 2}{6} = \frac{7}{6} = \alpha + \beta$

$\frac{3}{2} \times ( - \frac{1}{3} ) = - \frac{1}{2} = \alpha \beta$