anonymous
  • anonymous
find a, b, c if root of equation y=ax^2+bx+c root is (1/5)-(2/9)i i means imaginary
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Mertsj
  • Mertsj
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
anonymous
  • anonymous
Thanks but I'm having trouble multiplying it, but I understand what you mean, like the conjugates multiply to get the equation.
anonymous
  • anonymous
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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Mertsj
  • Mertsj
\[[x-(\frac{1}{5}-\frac{2}{9}i)][x-(\frac{1}{5}+\frac{2}{9}i)]=0\]
Mertsj
  • Mertsj
\[x ^{2}-(\frac{1}{5}+\frac{2}{9}i)x-(\frac{1}{5}-\frac{2}{9}i)x+(\frac{1}{5}-\frac{2}{9}i)(\frac{1}{5}+\frac{2}{9}i)\]
Mertsj
  • Mertsj
\[x ^{2}-\frac{1}{5}x-\frac{2}{9}ix-\frac{1}{5}x+\frac{2}{9}ix+\frac{1}{25}+\frac{4}{81}\]
Mertsj
  • Mertsj
\[x ^{2}-\frac{2}{5}x+\frac{181}{2025}=0\]
Mertsj
  • Mertsj
\[2025x ^{2}-810x+181=0\]
Mertsj
  • Mertsj
a= 2025, b=-810, c=181
Mertsj
  • Mertsj
Are you still with me?
anonymous
  • anonymous
Yup
Mertsj
  • Mertsj
Did I made any more mistakes?
anonymous
  • anonymous
That was a masterpiece
Mertsj
  • Mertsj
ty
anonymous
  • anonymous
no i think you did it perfectly
anonymous
  • anonymous
Thanks
Mertsj
  • Mertsj
yw
anonymous
  • anonymous
I just checked it and it worked :)
Mertsj
  • Mertsj
Great.

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