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
• (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

• Login to reply the answers
• 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

• Login to reply the answers
• 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.

• Login to reply the answers