Question

verify the relation between coefficient of quadratic polynomial 3x2+8x-3 with its zeroes​

Answer 1

Step-by-step explanation:

sum of zeroes= -b/a = -8/3

product of zerows= c/a= -3/

Answer 2

$\large\underline{\sf{Solution-}}$

Given quadratic polynomial is

$\rm :\longmapsto\:f(x) = {3x}^{2} + 8x - 3$

Let's first find the zeroes of quadratic polynomial.

$\rm :\longmapsto\:f(x) = {3x}^{2} + 9x - x - 3$

$\rm :\longmapsto\:f(x) = 3x(x + 3) - 1(x + 3)$

$\rm :\longmapsto\:f(x) = (x + 3)(3x - 1)$

So, to get zeroes of f(x),

$\rm :\longmapsto\:f(x) = (x + 3)(3x - 1) = 0$

$\rm\implies \:x = - 3 \: \: or \: \: x = \dfrac{1}{3}$

Let assume that

$\rm :\longmapsto\: \alpha = - 3$

and

$\rm :\longmapsto\: \beta = - \dfrac{1}{3}$

So,

$\rm :\longmapsto\:Sum \: of \: zeroes$

$\rm \:  =  \: \alpha + \beta$

$\rm \:  =  \: - 3 + \dfrac{1}{3}$

$\rm \:  =  \: \dfrac{ - 9 + 1}{3}$

$\rm \:  =  \: - \: \dfrac{8}{3}$

$\rm\implies \: \boxed{\tt{ \alpha + \beta   =  \: - \dfrac{8}{3} }}$

And

$\rm :\longmapsto\:Product \: of \: zeroes$

$\rm \:  =  \: \alpha \beta$

$\rm \:  =  \: - 3 \times \dfrac{1}{3}$

$\rm \:  =  \: - 1$

$\rm\implies \: \boxed{\tt{ \alpha \beta   =  \: - 1 }}$

VERIFICATION

Given quadratic polynomial is

$\rm :\longmapsto\:f(x) = {3x}^{2} + 8x - 3$

Now, We know that

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:Sum \: of \: zeroes = - \dfrac{8}{3}$

$\red{\rm\implies \:\boxed{\sf{ Sum \: of \: zeroes = \alpha + \beta \: }}}$

And, Also

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:Product \: of \: zeroes = \dfrac{ - 3}{3} = - 1$

$\red{\rm\implies \:\boxed{\sf{ Product \: of \: zeroes = \alpha \beta \: }}}$

HENCE, VERIFIED