Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

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?

Relevance
• 1 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

• 1 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

So:

a = 6, b = 19, c = 54, d = 90

• nbsale
Lv 6
1 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.