Question

find the zeroes of polynomial 3x²-x-4 and hence verify the relationship between coefficients and zeroes of polynomial​

Answer 1

Answer:

Let f(x) = 3x2 ˗ x ˗ 4 3x2 ˗ 4x + 3x ˗ 4 = x(3x ˗ 4) + 1 (3x ˗ 4) = (3x ˗ 4) (x + 1) To find the zeroes, set f(x) = 0 (3x ˗ 4) = 0 or (x + 1) = 0 x = 4/3 or x=-1 So, the zeroes of f(x) are 4/3 and x=-1 Again, Sum of zeroes = 4/3 + (-1) = 1/3 = -b/a = (-Coefficient of x)/(Cofficient of x2) Product of zeroes = 4/3 + (-1) = -4/3 = c/a = Constant term / Coefficient of x^2

Answer 2

Given :

→ 3x² - x - 4

To Find :

• The zeros of polynomial and verify the relationship between coefficients and zeros.

According to the question,

$\implies \sf 3x^2 - x - 4$

$\implies \sf 3x^2 - 4x + 3x - 4$

$\implies \sf x(3x - 4) + 1(3x - 4)$

$\implies \sf (x + 1) (3x - 4)$

$\therefore \bf x + 1 = 0$

$\implies \sf x = -1$

$\implies \sf 3x - 4 = 0$

$\implies \sf x = \dfrac{4}{3}$

VERIFICATION :-

Let α = - 1 and β = 4/3

$\\ \implies \sf \alpha + \beta = - 1 + \dfrac{4}{3} = \dfrac{1}{3} = \dfrac{ - 1}{3} = \dfrac{ - b}{a}$

$\\ \implies \sf \alpha \beta = - 1 \times \dfrac{4}{3} = \dfrac{- 4}{3} = \dfrac{c}{a}$

Hence, verified!

Know More :-

If alpha and beta are the zeros of the polynomial and a, b and c are the respective coefficients of the variables in the terms of the quadratic polynomial, then :-

$\sf \alpha + \beta = \dfrac{-b}{a}$

$\sf \alpha \times \beta = \dfrac{c}{a}$