Question

find the zeroes of the quardatic polynomials and verify the relationship between the zeroes and the cofficientsof 6x^2-3-7x

Answer 1

We have :

f(x) = 6x² - 7x - 3

= 6x² + 2x - 9x - 3

= 2x(3x-1) - 3(3x+ 1)

= (2x-3)(3x+1)

•°• f(x) = 0 , (2x-3)(3x+1)

=> 2x - 3 = 0. or 3x + 1 = 0

=> 2x = 3 or 3x = - 1

=> x = 3/2 or -1/3

So, the zeros of f(x) are 3/2 and -⅓.

Sum of Zeroes = -b/a => - (Coefficient of x)/Coefficient of x²

3/2 - 1/3 = 7/6

7/6 = 7/6


Now, Product of Zeros = c/a => Constant term/Coefficient of x²

3/2 * -1/3 = -3/6

-½ = -½


Hence , it's Verified!

Answer 2

f(x) = 6x² - 7x - 3

= 6x² + 2x - 9x - 3

= 2x(3x-1) - 3(3x+ 1)

= (2x-3)(3x+1)

= 2x - 3 = 0 and 3x + 1 = 0

= 2x = 3 or 3x = - 1

= x = 3/2 or -1/3

So, Sum of Zeroes = -b/a

= 3/2 - 1/3

= 7/6. ( Verified)

So, Product of Zeros = c/a

3/2 × -1/3

3/-6

-1/2

Verified