209k views
2 votes
Solve for the zeros of the quadratic function f(x) = 9x² + 6x +1. Write the answer as a fraction.

1.
2.
X=
x=
-b± √b²-4ac
2a
4. x=
-6±√(6)2-4(9)(1)
2(9)
3. x=-6±√36-36
18

User Callo
by
7.6k points

1 Answer

4 votes

Explanation:

To solve for the zeros of a quadratic function, we need to find the values of x where f(x) equals zero. This means that we need to set the quadratic function equal to zero and solve for x.

In this case, our quadratic function is f(x) = 9x² + 6x + 1. To set it equal to zero, we write:

9x² + 6x + 1 = 0

Now we can use the quadratic formula to solve for x:

x = (-b ± √b² - 4ac) / 2a

In this equation, a, b, and c are the coefficients of the quadratic function. In our case, a = 9, b = 6, and c = 1. Substituting these values into the equation, we get:

x = (-6 ± √36 - 4(9)(1)) / 2(9)

Simplifying, we get:

x = (-6 ± √0) / 18

Since the square root of 0 is 0, we have:

x = -6/18 and x = -6/18

Reducing the fractions, we get:

x = -1/3 and x = -1/3

Therefore, the zeros of the quadratic function f(x) = 9x² + 6x + 1 are x = -1/3 and x = -1/3.

User Wojciech Danilo
by
8.2k points