Aquadratic equation with real coefficients and leading coefficient 1, has x = -bi as a root. write the equation in general form.

x² + b²

Explanation:

1) Quadratic equation form, showing the two roots r₁ and r₂:

A (x - r₁)(x - r₂).

2) Coefficient 1 ⇒ A = 1

3) Complex roots ⇔ the roots are conjugate

4) r₁ = -bi ⇒ conjugate = bi = r₂

5) Replace -bi and bi for r₁ and r₂ in the general form:

[x - (-bi) ] [ x - bi] = (x + bi) (x - bi) = x² - (-bi)² = x² + b²

x² + b²

third option: 6x+8

step-by-step explanation:

we can see that the given question is a sum of series and the values of n vary from 0 to 3

the first term will be:

when n=0, [(0)*x+2] => (0+2) => 2

when n=1, [(1)*x+2] => (x+2) => x+2

when n=2, [(2)*x+2] => 2x+2

when n=4, [(3)*x+2] => 3x+2

so adding up all the terms will give us:

=2+x+2+2x+2+3x+2

=x+2x+3x+2+2+2+2

=6x+8

so,

third option is the correct answer ..

correct

step-by-step explanation: