Help to find the values of a quartic equation with i (imaginary)?
The equation is this: x^4 - ax^3 + bx^2 - cx + d = 0
I've already found the four roots (I was given two and simply found their conjugates). They are: 3i, -3i, 3 + i, and 3 - i.
I have to find the values of a and d (from the equation). I'm not sure how to go about this / find them. Help?
- Φ² = Φ+1Lv 71 month ago
(x-α)(x-β)(x-γ)(x-δ) = x⁴ - (α+β+γ+δ)x³ + (αβ+γδ+αγ+βδ+αδ+βγ)x² - (αβγ+αβδ+αγδ+βγδ)x + αβγδ
So a = 3i + -3i + (3 + i) + (3 - i) = 6
and d = (3i)(-3i)(3 + i)(3 - i) = 90
- llafferLv 71 month ago
If you know the roots, you can subtract them all from x, multiply the differences, set it equal to zero, and expand the left side to get your original polynomial:
roots are: ±3i and 3 ± i:
(x - 3i)(x + 3i)[x - (3 + i)][x - (3 - i)] = 0
(x - 3i)(x + 3i)(x - 3 - i)(x - 3 + i) = 0
Multiply the conjugate pairs:
(x² + 3xi - 3xi - 9i²)(x² - 3x + xi - 3x + 9 - 3i - xi + 3i - i²) = 0
Simplify. If you did it correctly all of the "i" terms should cancel out:
(x² - 9i²)(x² - 3x - 3x + 9 - i²) = 0
(x² - 9i²)(x² - 6x + 9 - i²) = 0
i² = -1, so:
(x² + 9)(x² - 6x + 9 + 1) = 0
(x² + 9)(x² - 6x + 10) = 0
Now multiply this and simplify:
x⁴ - 6x³ + 10x² + 9x² - 54x + 90 = 0
x⁴ - 6x³ + 19x² - 54x + 90 = 0
a = 6, b = 19, c = 54, d = 90
- nbsaleLv 61 month ago
Just multiply them out
(x-3i)(x+3i) = x^2 + 9 = A
(x - (3+i))(x - (3-i)) = x^2 -(3+i+3-i)x + (9+1))
= x^2 - 6x + 10 = B
Now just multiply A and B and you'll have the quartic.