Find the sum and product of zeroes of the quadratic polynomial 3x²-x-4​

2 thoughts on “Find the sum and product of zeroes of the quadratic polynomial 3x²-x-4​”

1. Given

   Roots of quadratic polynomials are [tex]\frac{4}{3},-1[/tex]

   explanation:

   Since we have given that

   [tex]3x^2-x-4[/tex]

   First we will find the zeroes of the quadratic polynomial.

   We will use “Split the middle terms”:

   [tex]3x^2-x-4=0\\\\3x^2+3x-4x-4=0\\\\3x(x+1)-4(x+1)=0\\\\(3x-4)(x+1)=0\\\\x=\frac{4}{3},-1[/tex]

   Now,

   Let, [tex]\alpha =\frac{4}{3},\beta =-1[/tex]

   Now, we will verify the relationship between the zeroes and coefficient.

   Sum of zeroes is given by

   [tex]\alpha +\beta =\frac{4}{3}-1=\frac{1}{3}\\\\\alpha \beta =-1\times \frac{4}{3}=\frac{-4}{3}\\and\\\\\alpha +\beta =\frac{-b}{a}=\frac{1}{3},\alpha\beta =\frac{c}{a}=\frac{-4}{3}[/tex]

   Hence, verified.

   Roots of quadratic polynomials are [tex]\frac{4}{3},-1[/tex]

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2. Answer:

   By splitting the middle term,

   3x²-x-4

   = 3x²+3x-4x-4

   = 3x(x+1)-4(x+1)

   = (3x-4)(x+1)

   So, the zeroes of the polynomial are x = 4/3 and x = -1.

   Sum = 4/3 + (-1) = 4/3 – 1 = 1/3

   Product = 4/3 × (-1) = -4/3

   Reply