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

1. 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

2. anonymous

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

3. 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

4. Mertsj

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

5. Mertsj

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

6. Mertsj

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

7. Mertsj

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

8. Mertsj

$2025x ^{2}-810x+181=0$

9. Mertsj

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

10. Mertsj

Are you still with me?

11. anonymous

Yup

12. Mertsj

Did I made any more mistakes?

13. anonymous

That was a masterpiece

14. Mertsj

ty

15. anonymous

no i think you did it perfectly

16. anonymous

Thanks

17. Mertsj

yw

18. anonymous

I just checked it and it worked :)

19. Mertsj

Great.