Use the quadratic formula to complete the table. 3x^2+4x+4=0
3x^2+2x+4=0
9x^2-6x+2=0
Value of Discriminant and Solutions

For 3x^2+4x+4=0

Discriminant= = -32

The solutions are

(-b+√x)/2a= (-2+2√-2)/3

(-b-√x)/2a= (-2-2√-2)/3

For 3x^2+2x+4=0

Discriminant= -44

The solutions

(-b+√x)/2a= (-1+√-11)/3

(-b-√x)/2a= (-1-√-11)/3

For 9x^2-6x+2=0

Discriminant= -36

The solutions

(-b+√x)/2a= (1+√-1)/3

(-b-√x)/2a= (1-√-1)/3

Step-by-step explanation:

Formula for the discriminant = b²-4ac

let the discriminant be = x for the equations

The solution of the equations

= (-b+√x)/2a and = (-b-√x)/2a

For 3x^2+4x+4=0

Discriminant= 4²-4(3)(4)

Discriminant= 16-48

Discriminant= = -32

The solutions

(-b+√x)/2a =( -4+√-32)/6

(-b+√x)/2a= (-4 +4√-2)/6

(-b+√x)/2a= (-2+2√-2)/3

(-b-√x)/2a =( -4-√-32)/6

(-b-√x)/2a= (-4 -4√-2)/6

(-b-√x)/2a= (-2-2√-2)/3

For 3x^2+2x+4=0

Discriminant= 2²-4(3)(4)

Discriminant= 4-48

Discriminant= -44

The solutions

(-b+√x)/2a =( -2+√-44)/6

(-b+√x)/2a= (-2 +2√-11)/6

(-b+√x)/2a= (-1+√-11)/3

(-b-√x)/2a =( -2-√-44)/6

(-b-√x)/2a= (-2 -2√-11)/6

(-b-√x)/2a= (-1-√-11)/3

For 9x^2-6x+2=0

Discriminant= (-6)²-4(9)(2)

Discriminant= 36 -72

Discriminant= -36

The solutions

(-b+√x)/2a =( 6+√-36)/18

(-b+√x)/2a= (6 +6√-1)/18

(-b+√x)/2a= (1+√-1)/3

(-b-√x)/2a =( 6-√-36)/18

(-b-√x)/2a= (6 -6√-1)/18

(-b-√x)/2a= (1-√-1)/3

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source