6x^2-3-7
6x^2-7x-3
6x^2-9x+2x-3
3x(2x-3x)+1(2x-3)
(2x-3)(3x+1)
2x-3=0. 3x+1=0
x =3/2. x=-1/3
a=6
b=-7
c=-3
Answers
the anwer is 6 is the correct answer
Answer:
Step-by-step explanation:
1. Given f(x) = 6x2 – 7x – 3To find the zerosLet 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 = 0x = 3/2 x = -1/3It 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 verificationSum of zeros = – coefficient of x / coefficient of x23/2 + (-1/3) = – (-7) / 6 7/6 = 7/6 Product of roots = constant / coefficient of x23/2 x (-1/3) = (-3) / 6 -1/2 = -1/2Therefore, the relationship between zeros and their coefficients is verified.2.Let f(x) = 3x2 ˗ x ˗ 4 = 03x2 ˗ 4x + 3x ˗ 4 = 0⇒ x(3x ˗ 4) + 1 (3x ˗ 4) = 0⇒ (3x ˗ 4) (x + 1) = 0To find the zeroes(3x ˗ 4) = 0 or (x + 1) = 0x = 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