Asked by Lina
Which equation has the complex number 4-3i as a root?
1) x^2+6x-25=0
2) x^2-6x+25=0
3) x^2+8x-25=0
4) x^2-8x+25=0
Can anyone please teach me how to do it?THANKS A LOT!!!
1) x^2+6x-25=0
2) x^2-6x+25=0
3) x^2+8x-25=0
4) x^2-8x+25=0
Can anyone please teach me how to do it?THANKS A LOT!!!
Answers
Reiny
if 4-3i is a root, so is 4+3i
then factors are (x-4+3i) and (x-4-3i)
set (x-4+3i)(x-4-3i) = 0 and expand to see which of the equation you get.
(remember that i^2 = -1)
then factors are (x-4+3i) and (x-4-3i)
set (x-4+3i)(x-4-3i) = 0 and expand to see which of the equation you get.
(remember that i^2 = -1)
Lina
y the factors are (x-4+3i) and (x-4-3i) ???Can you explain please?