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# select the approxiate value of x that is solutions for (x)=0 value f(x)=-3x^2+2x+8

select the approxiate value of x that is solutions for (x)=0 value f(x)=-3x^2+2x+8

Find the appropriate value(s) of x that is/are the solution(s) to f(x)=0 where f(x)=-3x^2+2x+8.

The values of x that make f(x) = 0 are called the Zeros of f(x). The Zeros are also the x-intercepts if they are real. If the Zeros are Imaginary then f(x) has no x-intercepts.

So we want to solve the Quadratic Equation 0=-3x^2+2x+8 for the x-values that make it true, its Zeros, a.k.a., Roots.

Let's use the Quadratic Formula to find the Zeros:

If 0 = ax^2 + bx + c, then
x = (- b ± sqrt(b^2-4ac))/(2a)

Identify parameters:
a = -3, the coefficient of x^2
b = 2, the coefficient of x
c = 8, the constant

Substitute:
x = (- 2 ± sqrt(2^2-4(-3)(8)))/(2(-3))

Simplify:
x = (- 2 ± sqrt(4+4(24)))/(-2(3))
x = (- 2 ± sqrt(4(1+24)))/(-2(3))
x = (- 2 ± sqrt(4(25)))/(-2(3))
x = (- 2 ± 2*5)/(-2(3))
x = (1 ± 5)/3
x = -4/3 or 2
Hi again Michelle;
0=f(x)=-3x2+2x+8
For the FOIL...
FIRST must be (3x)(x) and one number must be negative to produce -3x2.
OUTER and INNER must add-up to 2x.
LAST must be (8)(1) or (1)(8) or (4)(2) or (2)(4) and both numbers must be positive to produce positive 8, as well as positive 2x for the OUTER and INNER.
0=(3x+4)(-x+2)
Let's FOIL...
FIRST...(3x)(-x)=-3x2
OUTER...(3x)(2)=6x
INNER...(4)(-x)=-4x
LAST...(4)(2)=8
0=-3x2+6x-4x+8
0=-3x2+2x+8
(3x+4)(-x+2)=0
Either or both parenthetical equations must equal zero.
3x+4=0
3x=-4
x=-4/3

-x+2=0
-x=-2
x=2