F(x) = ax²+bx+c
The above function, f(x), has a quadratic equation, where a, b, and c are constants. If the sum of the roots is -2/3 and the product of the roots is 1/6, then which of the following represents the quadratic equation, f(x)?
(a) f(x)= 6x²+4x+6
(b) f(x)= 1/6x²+4x+1
(c) f(x)= 3x²-2x+1
(d) f(x)= 6x²+4x+1
To find the quadratic equation, we need to find the values of a, b, and c.
The sum of the roots is -2/3, which means that the equation can be written as:
(x - r1)(x - r2) = 0
Where r1 and r2 are the roots. Expanding this equation, we get:
x² - (r1 + r2)x + r1r2 = 0
Comparing this with the quadratic equation f(x) = ax² + bx + c, we can see that a = 1, b = -(r1 + r2), and c = r1r2.
The product of the roots is 1/6, so we have r1r2 = 1/6.
Substituting r1r2 = 1/6 into the equation a = 1, b = -(r1 + r2), and c = r1r2, we get:
a = 1
b = -(-2/3) = 2/3
c = 1/6
So the quadratic equation f(x) is:
f(x) = x² + (2/3)x + 1/6
Multiplying both sides by 6 to eliminate the fractions, we get:
6f(x) = 6x² + 4x + 1
Therefore, the correct answer is (d) f(x) = 6x² + 4x + 1.