Question

Find the zeros of the quadratic polynomial (6x2-7x-3) and verify the relation between its zeros and coefficients.

Answer 1

$6 {x}^{2} - 7x - 3 = 0 \\ 6 {x}^{2} - 9x + 2x - 3 = 0 \\ 3x(2x - 3) + 1(2x - 3) \\( 3x + 1)(2x - 3) = 0 \\ 3x + 1 = 0 \\ 3x = - 1 \\ x = \frac{ - 1}{3} \\ \: and \\ \: 2x - 3 \: = 0 \\8 2x = 3 \\x = \: \frac{3}{2} \\ sum \: of \: zeroes \: = \alpha + \beta = \frac{ - 1}{3} + \frac{3}{2} = \frac{7}{6} = - \frac{coefficient \: of \: x}{coefficient \: of \: x^{2} } \\ product \: \: ofzeroes = \alpha \beta = \frac{ - 1}{3} \times \frac{3}{2} = \frac{ - 1}{ 2 } = \frac{constant \: term}{coefficient \: of \: x^{2} }$

Answer 2

Hey Mate! Here is Ur Answer...