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anonymous
 4 years ago
find a, b, c if root of equation y=ax^2+bx+c
root is (1/5)(2/9)i
i means imaginary
anonymous
 4 years ago
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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Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1If 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][x1/5+(2/9)i]=0 and simplify into the ax^2+bx+c form

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks but I'm having trouble multiplying it, but I understand what you mean, like the conjugates multiply to get the equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Also, 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

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1\[[x(\frac{1}{5}\frac{2}{9}i)][x(\frac{1}{5}+\frac{2}{9}i)]=0\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1\[x ^{2}(\frac{1}{5}+\frac{2}{9}i)x(\frac{1}{5}\frac{2}{9}i)x+(\frac{1}{5}\frac{2}{9}i)(\frac{1}{5}+\frac{2}{9}i)\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1\[x ^{2}\frac{1}{5}x\frac{2}{9}ix\frac{1}{5}x+\frac{2}{9}ix+\frac{1}{25}+\frac{4}{81}\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1\[x ^{2}\frac{2}{5}x+\frac{181}{2025}=0\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1\[2025x ^{2}810x+181=0\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1Did I made any more mistakes?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That was a masterpiece

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i think you did it perfectly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just checked it and it worked :)
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