Question

find a quadratic equation with the roots -1+4i and -1-4i

Answer 1

quadratic equation with roots x=a and x=b are given by (x-a)(x-b)=0

Answer 2

so we have (x-(-1+4i))(x-(-1-4i))=0

Answer 3

implies (x+1-4i)(x+1+4i)=0
implies (x+1)^2+16=0

Answer 4

implies x^2+2x+17=0

Answer 5

@Hope214

Answer 6

i dont understand can you please explain?

Answer 7

ok.. see if the roots are a and b, the equation is given by (x-a)(x-b)=0

Answer 8

here a=-1+4i and b=-1-4i so plug them in the equation (x-a)(x-b)=0

Answer 9

ok so once you plug them in what do you do

Answer 10

then we reduce it in the general quadratic form which is px^2+qx+r=0 where p,q,r are constants

Answer 11

now lets get back to the problem, after substitutiong we have (x-(-1+4i))(x-(-1-4i))=0
implies (x+1-4i)(x+1+4i)=0

Answer 12

implies {(x+1)-4i}{(x+1)+4i}=0 now use the formula (a-b)(a+b)=a^2-b^2
and you get (x+1)^2-(4i)^2=0
implies (x+1)^2+16=0

Answer 13

implies x^2+2x+1+16=0
implies x^2+2x+17=0

Answer 14

ok thank u