Find the zeros:
f(x)=x^2 +x-20
which two factors of 20 differ by 1?
Those will be the zeros; one plus, one minus.
F(x) = x^2 + x - 20.
x^2 + x - 20 = 0,
(x-4)(x+5) = 0,
x-4 = 0. X =
x + 5 = 0. X =
To find the zeros of the quadratic function f(x) = x^2 + x - 20, we need to determine the values of x where the function equals zero. In other words, we are looking for the x-values that make f(x) = 0.
To solve for the zeros, we can set f(x) = 0 and use factoring, completing the square, or the quadratic formula.
1. Factoring:
Rearrange the equation to f(x) = x^2 + x - 20 = 0. Looking at the expression, we want to find two numbers that multiply to give -20 and add up to 1 (the coefficient of x).
The two numbers that satisfy this condition are 5 and -4. Rewriting the equation using these numbers, we have:
(x + 5)(x - 4) = 0
Now, set each factor equal to zero and solve for x:
x + 5 = 0 => x = -5
x - 4 = 0 => x = 4
So, the zeros of the function f(x) = x^2 + x - 20 are x = -5 and x = 4.
2. Completing the Square:
Rearrange the equation to f(x) = x^2 + x = 20. To complete the square, we need to add and subtract the constant term that makes the quadratic expression a perfect square trinomial.
Start by taking half the coefficient of x, which is 1/2 (or 0.5 in decimal form). Then square this value to get 1/4 (or 0.25). Add and subtract 0.25 to the equation:
x^2 + x + 0.25 - 0.25 = 20
Rearrange and factor the perfect square trinomial:
(x + 0.5)^2 - 0.25 = 20
Now, isolate the perfect square term:
(x + 0.5)^2 = 20 + 0.25
(x + 0.5)^2 = 20.25
Take the square root of both sides:
x + 0.5 = ±√20.25
x + 0.5 = ±4.5
Solve for x:
x = -0.5 ± 4.5
This gives us two solutions:
x = -0.5 + 4.5 = 4
x = -0.5 - 4.5 = -5
Therefore, the zeros of the function f(x) = x^2 + x - 20 are x = -5 and x = 4.
3. Quadratic Formula:
The quadratic formula can be used to find the zeros of any quadratic function. For the equation f(x) = x^2 + x - 20 = 0, we have:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 1, and c = -20. Substituting these values into the quadratic formula, we get:
x = (-(1) ± √((1)^2 - 4(1)(-20))) / (2(1))
x = (-1 ± √(1 + 80)) / 2
x = (-1 ± √81) / 2
x = (-1 ± 9) / 2
This gives two solutions:
x = (-1 + 9) / 2 = 8 / 2 = 4
x = (-1 - 9) / 2 = -10 / 2 = -5
Hence, the zeros of the function f(x) = x^2 + x - 20 are x = -5 and x = 4.