Answer: 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