Question

Assume that 2-3i is a solution of ax^2+bx+c=0, where a,b,c are real numbers, which of the following is also a solution of the equation? show WORK!!!

2+3i
-2-3i
-2+3i
3+2i
1/(2-3i)
Assume that 2-3i is a solution of ax^2+bx+c=0, where a,b,c are real numbers, which of the following is also a solution of the equation? show WORK!!!

2+3i
-2-3i
-2+3i
3+2i
1/(2-3i)
@Mathematics

Answer 1

@curry this is not any work to show. if
\[a+bi\] is a zero of a polynomial with real coefficients then so is
\[a-bi\]

Answer 2

^

Answer 3

i guess we could say we know this by the quadratic formula. it says the solutions are
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] and so if
\[b^2-4ac<0\] you have two complex solutions,
\[-\frac{b}{2a}+\frac{\sqrt{4ac-b^2}}{2a}i\] and
\[-\frac{b}{2a}-\frac{\sqrt{4ac-b^2}}{2a}i\]

Answer 4

you can actually show it though using the formula but it's implied

Answer 5

i wonder if you could solve it by = the one to your answer and then finding a b and c lol

Answer 6

by using like another formula