To find the zeros of the quadratic function f(x) = 6x² + 12x – 7, we need to solve the equation f(x) = 0. This leads us to:
6x² + 12x – 7 = 0
We can use the quadratic formula, which is:
x = (-b ± √(b² – 4ac)) / (2a)
In our quadratic equation, the coefficients are:
- a = 6
- b = 12
- c = -7
First, we need to calculate the discriminant (b² – 4ac):
b² = 12² = 144
4ac = 4 * 6 * (-7) = -168
Now, let’s find the discriminant:
144 – (-168) = 144 + 168 = 312
Since the discriminant is positive, we will have two distinct real roots. Now, we can plug the values back into the quadratic formula:
x = (-12 ± √312) / (2 * 6)
Calculating further:
x = (-12 ± √312) / 12
√312 = √(4 * 78) = 2√78
Thus, the equation simplifies to:
x = (-12 ± 2√78) / 12
Breaking this down gives us:
x = -1 ± (√78 / 6)
Therefore, the two zeros of the quadratic function are:
x₁ = -1 + (√78 / 6)
x₂ = -1 – (√78 / 6)
In decimal form, you can approximate the values of these zeros using a calculator:
- x₁ ≈ 0.354
- x₂ ≈ -2.354
And there you have it! The zeros of the quadratic function f(x) = 6x² + 12x – 7 are approximately 0.354 and -2.354.