Question

find a, b, c if root of equation y=ax^2+bx+c
root is (1/5)-(2/9)i

i means imaginary

Answer 1

If 1/5 -(2/9)i is a root then 1/5+(2/9)i is also a root. So write that
[x-(1/5-(2/9)i][x-1/5+(2/9)i]=0 and simplify into the ax^2+bx+c form

Answer 2

Thanks but I'm having trouble multiplying it, but I understand what you mean, like the conjugates multiply to get the equation.

Answer 3

Also, aren't you supposed to switch the + and - in the second part?
[x-(1/5-(2/9)i][x+1/5-(2/9)i]=0

Answer 4

\[[x-(\frac{1}{5}-\frac{2}{9}i)][x-(\frac{1}{5}+\frac{2}{9}i)]=0\]

Answer 5

\[x ^{2}-(\frac{1}{5}+\frac{2}{9}i)x-(\frac{1}{5}-\frac{2}{9}i)x+(\frac{1}{5}-\frac{2}{9}i)(\frac{1}{5}+\frac{2}{9}i)\]

Answer 6

\[x ^{2}-\frac{1}{5}x-\frac{2}{9}ix-\frac{1}{5}x+\frac{2}{9}ix+\frac{1}{25}+\frac{4}{81}\]

Answer 7

\[x ^{2}-\frac{2}{5}x+\frac{181}{2025}=0\]

Answer 8

\[2025x ^{2}-810x+181=0\]

Answer 9

a= 2025, b=-810, c=181

Answer 10

Are you still with me?

Answer 11

Yup

Answer 12

Did I made any more mistakes?

Answer 13

That was a masterpiece

Answer 14

ty

Answer 15

no i think you did it perfectly

Answer 16

Thanks

Answer 17

yw

Answer 18

I just checked it and it worked :)

Answer 19

Great.