To find the zeros of the quadratic function f(x) = 2x² + 10x + 3, we need to determine the values of x for which f(x) = 0. The zeros can be found using the quadratic formula:
x = (-b ± √(b² – 4ac)) / 2a
In our case, the coefficients are:
- a = 2
- b = 10
- c = 3
First, we calculate the discriminant (b² – 4ac):
b² – 4ac = 10² – 4(2)(3) = 100 – 24 = 76
Since the discriminant is positive, we can expect two distinct real roots. Now, we can substitute the values of a, b, and the discriminant back into the quadratic formula:
x = (-10 ± √76) / (2 * 2)
Calculating further gives us:
x = (-10 ± √76) / 4
The square root of 76 simplifies to:
√76 = 2√19
Now substituting back, we get:
x = (-10 ± 2√19) / 4
This further simplifies to:
x = (-5 ± √19) / 2
Thus, the two zeros of the quadratic function are:
- x = (-5 + √19) / 2
- x = (-5 – √19) / 2
These values represent the points at which the graph of the quadratic function intersects the x-axis. In summary, the zeros are:
- x ≈ -0.9
- x ≈ -4.1