Question

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

Answer 1

Answer:

$let \: \: p(x) = 6x {}^{2} - 7x - 3 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6x {}^{2} + ( - 9 + 2)x - 3 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6x {}^{2} - 9x + 2x - 3 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = 3x(2x - 3) + 1(2x - 3) \\ \: \: \: \: \: \: \: \: \: \: \: =( 3x + 1)(2x - 3) \\ for \: zeros \: \\ p(x) = 0 \\(3x + 1) = 0 \\ 3x = - 1 \\ x = \frac{ - 1}{3} \: \: and \: \\ (2x - 3) = 0 \\2 x = 3 \\ x = \frac{3}{2} \\ let \: \alpha \: and \: \beta are \: the \: zeros \: of \: p(x) \\ then \: \\ \alpha = \frac{ - 1}{3} \: and \: \beta = \frac{3}{2} \\ sum \: of \: zeros \: = \alpha + \beta \\ \frac{ - b}{a} = \frac{ - 1}{3} + \frac{3}{2} = \frac{ - 2 + 9}{6} = \frac{ 7}{6} \\ product \: of \: zeros \: = \alpha \beta \\ \frac{c}{a} = \frac{ - 1}{3} \times \frac{3}{2} = \frac{ - 3}{6}$

Answer 2

Answer:

Hey mate your answer is given below