Complex Numbers & Quadratic Equations Test

Result

Complex Numbers & Quadratic Equations Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

If the cube roots of unity are 1, ω, ω² then the roots of the equation
(x – 1)³ + 8 = 0, are

A
-1 , - 1 + 2ω, - 1 - 2ω²

B
-1 , -1, - 1

C
-1 , 1 - 2ω, 1 - 2ω²

D
-1 , 1 + 2ω, 1 + 2ω²

Solution

(x – 1)³ + 8 = 0
⇒ (x – 1) = (-2) (1)^1/3
⇒ x – 1 = -2 or -2ω or -2ω²
or n = -1 or 1 – 2ω or 1 – 2ω²
Hence, option C is correct.
Question 2
1 / -0

Let z₁ and z₂ be two roots of the equation z² + az + b = 0, z being complex. Further, assume that the origin, z₁ and z₂ form an equilateral triangle, then

A
a² = b

B
a² = 2b

C
a² = 3b

D
a² = 4b

Solution

Given that z₁ and z₂ be two roots of the equation z²+ az + b = 0, where z is a complex number.
So, we have z₁ + z₂ = -a, z₁z₂ = b (sum and product of roots)
Since z₁ and z₂form an equilateral triangle with the origin, we have

z₂ = z₁ (cos 60° + isin 60°)
= z₁ (1/2 + i√3/2)
Or, 2z₂ – z₁ = √3 i z₁
This gives, (2z₂ -z₁)² = -3z₁²
Hence, (z₁² + z₂²) = z₁z₂
So, a² – 2b = b
this gives a² = 3b.