What are the zeroes of f(x) = 2x^2 -3x + 1
x=1
x=1/2
I use the quadratic formula.
To find the zeroes of the quadratic function f(x) = 2x^2 - 3x + 1, we need to solve for x when f(x) equals zero.
Step 1: Set f(x) = 0:
2x^2 - 3x + 1 = 0
Step 2: Factor or apply the quadratic formula to solve for x:
Since the quadratic equation does not easily factor, we can use the quadratic formula to find the zeroes.
The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our quadratic equation of 2x^2 - 3x + 1 = 0, the coefficients are:
a = 2, b = -3, and c = 1.
Plugging these values into the formula, we get:
x = (-(-3) ± √((-3)^2 - 4(2)(1))) / (2(2))
x = (3 ± √(9 - 8)) / 4
x = (3 ± √1) / 4
x = (3 ± 1) / 4
Thus, we have two solutions for x:
x1 = (3 + 1) / 4 = 4 / 4 = 1
x2 = (3 - 1) / 4 = 2 / 4 = 1/2
Therefore, the zeroes of the quadratic function f(x) = 2x^2 - 3x + 1 are x = 1 and x = 1/2.