Use the quadratic formula to find the zeros of f
F(x)=2x^2-4x-11
{ 4 +/- sqrt (16 + 88) ] /4
[ 4 +/- sqrt(104) ]/ 4
[ 4 +/- 2 sqrt (26) ]/4
1 +/- 0.5 sqrt(26)
Wouldn't it be 16-88?
Oh, wait nvm. I didn't notice the negative sign on the 11
To find the zeros of the quadratic function f(x) = 2x^2 - 4x - 11, we can use the quadratic formula. The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this formula, a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
For the given quadratic function f(x) = 2x^2 - 4x - 11, we can identify the following values:
a = 2
b = -4
c = -11
Substituting these values into the quadratic formula, we have:
x = (-(-4) ± √((-4)^2 - 4(2)(-11))) / (2(2))
Simplifying further:
x = (4 ± √(16 + 88)) / 4
x = (4 ± √104) / 4
x = (4 ± 2√26) / 4
Now, we can simplify the expression inside the square root:
x = (4 ± 2√26) / 4
x = 2 ± √26 / 2
To find the zeros of the quadratic function f(x), we set f(x) equal to zero and solve for x. The zeros occur when the function intersects the x-axis.
Setting f(x) = 0, we have:
2x^2 - 4x - 11 = 0
Using the quadratic formula, we've found that the zeros of the given quadratic function f(x) = 2x^2 - 4x - 11 are:
x = 2 + √26 / 2
x = 2 - √26 / 2