Question

Find the zeros of the polynomial 6x2 -3-7x and verify the
relationship between the zeros and the coefficients. (CBSE 2008)
​

Answer 1

Step-by-step explanation:

p(x) = 6x²-3-7x

= 6x²-7x-3

= 6x²-9x+2x-3

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

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

For p(x) = 0

Either 2x-3= 0 => x= 3/2

or, 3x+1 =0 => x= -1/3

the zeros are 3/2, and -1/3

Sum of zeros= -b/a

3/2+(-1/3)=-(-7)/6

9-2/6=7/6

LHS=RHS

Product of zeros= c/a

3/2×-1/3=-3/6

-1/2=-1/2

LHS=RHS

Relationship verified.

Thanks

Answer 2

given,

6x²-7x - 3

to find the zeroes of the polynomial,

6x²-7x - 3 = 0

6x²-9x +2x -3 = 0

3x(2x-3) + 1(2x -3) = 0

(3x+1) (2x-3) = 0

x = -1/3 or 3/2

sum of the zeroes = 3/2 - 1/3 = 7/6 = - ( co-efficient of x) / co-efficient of x²

product of the zeroes = -1/3 × 3/2 = -1/2  = constant / co-efficient of x²