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

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

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

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Other answers:

\[[x-(\frac{1}{5}-\frac{2}{9}i)][x-(\frac{1}{5}+\frac{2}{9}i)]=0\]
\[x ^{2}-(\frac{1}{5}+\frac{2}{9}i)x-(\frac{1}{5}-\frac{2}{9}i)x+(\frac{1}{5}-\frac{2}{9}i)(\frac{1}{5}+\frac{2}{9}i)\]
\[x ^{2}-\frac{1}{5}x-\frac{2}{9}ix-\frac{1}{5}x+\frac{2}{9}ix+\frac{1}{25}+\frac{4}{81}\]
\[x ^{2}-\frac{2}{5}x+\frac{181}{2025}=0\]
\[2025x ^{2}-810x+181=0\]
a= 2025, b=-810, c=181
Are you still with me?
Yup
Did I made any more mistakes?
That was a masterpiece
ty
no i think you did it perfectly
Thanks
yw
I just checked it and it worked :)
Great.

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