Find the zeroes of the polynomial and verify the relationship between the zeroes and coefficients 1. 6x^2 - 3 - 7x
2. 3x^2 - x - 4

Step-by-step explanation:

1. Given  f(x) = 6x2 – 7x – 3

To find the zeros

Let us put f(x) = 0

⇒ 6x2 – 7x – 3 = 0

⇒ 6x2 – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (2x – 3)(3x + 1) = 0

⇒ 2x – 3 = 0 Or 3x + 1 = 0

x = 3/2                      x = -1/3

It gives us 2 zeros, for x = 3/2 and x = -1/3  

Hence, the zeros of the quadratic equation are 3/2 and -1/3.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

3/2 + (-1/3) = – (-7) / 6 7/6 = 7/6  

Product of roots = constant / coefficient of x2

3/2 x (-1/3) = (-3) / 6 -1/2 = -1/2

Therefore, the relationship between zeros and their coefficients is verified.

2.Let f(x) = 3x2 ˗ x ˗ 4 = 0

3x2 ˗ 4x + 3x ˗ 4  = 0

⇒ x(3x ˗ 4) + 1 (3x ˗ 4)  = 0

⇒  (3x ˗ 4) (x + 1) = 0

To find the zeroes

(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  x (-1) = -4/3 = c/a=

= Constant term / Coefficient of x2

Therefore, the relationship between zeros

a=r

d=q

c=p

b=s

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