Question

find the zeros of the quadratic polynomial 6x square - 7 x minus 3 ,and verify the relationship between the zeros and the co-efficients​

Answer 1

Step-by-step explanation:

6xsquare-7x-3

6x square -9x+2x -3

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

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

3x+1=0

x= -1/3

2x-3=0

x=3/2

Answer 2

Answer:

Zeroes of 6$x^{2}$ - 7x - 3 are 3/2 and -1/3.

Step-by-step explanation:

6$x^{2}$ - 7x - 3

6$x^{2}$ - 9x + 2x - 3

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

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

2x - 3 = 0

x = 3/2

3x + 1 = 0

x = -1/3

Verification :

Sum of zeroes = - ( coefficient of x )/coefficient of $x^{2}$

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

                   = 9 - 2/6

                   = 7/6

Product of zeroes = constant term/coefficient of $x^{2}$

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

Verified.

I hope this answer is helpful for you.