Question

Obtain the zeros of the quadratic polynomial 6x2 -3 -7x and verify the relation between its zeros and coefficients.​

Answer 1

Given,

$\sf f(x) = 6x^2 - 3 - 7x$

First we have to write in standard form,

$\sf 6x^2 - 7x - 3= 0$

$\sf 6x^2 - 9x + 2x - 3= 0$

$\sf 3x(2x - 3)+ 2x - 3= 0$

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

$\sf 3x + 1 = 0 \: and \: 2x - 3= 0$

$\sf x = \dfrac{ - 1}{3} \: and \: x= \dfrac{3}{2}$

The zeros of f(x) are 3/2 and -1/3.

We know that,

$\sf \alpha + \beta = \dfrac{ - b}{a} = \dfrac{ - (coefficient \: of \:x )}{coefficient \: of \: {x}^{2} }$

$\sf sum \: of \: zeros = \dfrac{3}{2} + \dfrac{ - 1}{3}$

$\sf = \dfrac{3}{2} - \dfrac{1}{3}$

$\sf = \dfrac{9 - 2}{6}$

$\sf = \dfrac{7}{6}$

$\sf \alpha \beta = \dfrac{c}{a} = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

$\sf product \: of \: zeros = \dfrac{3}{2} \times \dfrac{ - 1}{3}$

$\sf = \dfrac{ - 3}{6}$