Question

Find the zeroes of the polynomials and verify the relationship between zeroes and co-efficients ....
$6x { }^ {2} - 3 - 7x$

Answer 1

hey mate
here's the solution

Answer 2

$6 {x}^{2} - 3 - 7x \\ \\ 6 {x}^{2} - 7x - 3 \\ \\ 6 {x}^{2} - 9x + 2x - 3 \\ \\ 3x(2x - 3) + 1(2x - 3) \\ \\ (3x + 1)(2x - 3)$
To find zeros,put factors equal to zero

$3x + 1 = 0 \\ \\ x = \frac{ - 1}{3} \\ \\ 2x - 3 = 0 \\ \\ x = \frac{3}{2}$
Sum of zeros :
$= - \frac{1}{3} + \frac{3}{2} \\ \\ \frac{ - 2 + 9}{6} \\ \\ = \frac{7}{6} ...eq1$

product of zeros:

$\frac{ - 1}{3} \times \frac{3}{2} \\ \\ = - \frac{1}{2} ....eq2$
Sum of zeros as coefficient:

Sum of zeros =
$\frac{ - b}{a} \\ \\ = - \frac{ - 7}{6} \\ \\ = \frac{7}{6} ..eq3$
product of zeros as coefficient
$= \frac{c}{a} \\ \\ = \frac{ - 3}{6} \\ \\ = \frac{ - 1}{2} ...eq4$
you can see eq1 = eq3,
eq2= eq4

Hope it helps you