Question

Complex Roots

What are the roots of the equation x^2 - 12x + 40 = 0 in simplest a + bi form?

Answer 1

x= 6 +/- 2i

Explanation:

+/- is the positive or negative symbol

formula: ax^2+bx+c=0
And: X^2-12x+40=0

Then: a= 1 b= -12 c= 1

because none of the greatest common factors will factor into the problem, use the quadratic formula

x= -b +/- sqrt(b^2 -4ac) / 2a

Plug in the numbers

x= -(-12) +/- sqrt[ (-12)^2 -4(1)(40) ] / 2(1)

-(-12)=12

(-12)^2=144

(-4)(1)(40)=-160

x= 12 +/- sqrt(144-160) / 2

-160+144= -16

x= 12 +/- sqrt(-16) / 2

keep in mind

sqrt(-1) = i

i^2 = -1

sqrt(16)=4

so,

x= 12 +/- 4i / 2

Take out the 2’s from numerator (essentially dividing)

x= 2 (6 +/- 2i)

————————-

2

The 2’s in the numerator (outside of the parentheses) and in the denominator cancel each other out

Leaving the answer being

x= 6 +/- 2i