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If the real roots of the equation 3x4 – 6x3 3x2 – 54x – 216 = 0 are –2 and 4, what are the nonreal roots?

User ZeWaren
by
6.8k points

1 Answer

3 votes

Answer:

x = 4 or x = -2 or x = 3 i or x = -3 i

Explanation:

Solve for x:

3 (x^4 - 2 x^3 + x^2 - 18 x - 72) = 0

Divide both sides by 3:

x^4 - 2 x^3 + x^2 - 18 x - 72 = 0

The left-hand side factors into a product with three terms:

(x - 4) (x + 2) (x^2 + 9) = 0

Split into three equations:

x - 4 = 0 or x + 2 = 0 or x^2 + 9 = 0

Add 4 to both sides:

x = 4 or x + 2 = 0 or x^2 + 9 = 0

Subtract 2 from both sides:

x = 4 or x = -2 or x^2 + 9 = 0

Subtract 9 from both sides:

x = 4 or x = -2 or x^2 = -9

Take the square root of both sides:

Answer: x = 4 or x = -2 or x = 3 i or x = -3 i

PS: next time, please present the equation properly and do not make us guess.

User Florie
by
6.7k points
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