# Let z 1 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 ?

Grade:11

## 3 Answers

Latika Leekha
askIITians Faculty 165 Points
8 years ago
Hello student,
Given that z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex.
So, we have z1 + z2 = -a, z1z2 = b (sum and product of roots)
So, z2 = z1 ( cos 60° + i sin 60°)
= z1 (1/2 + √3/2 i)
So, 2z2 – z1 = √3 i z1
This gives, (2z2 -z1)2 = -3z12
Hence, (z12 + z22) = z1z2
So, a2 – 2b = b
this gives a2 = 3b.
Anupam Verma
26 Points
6 years ago
As z1, z2 are roots of z2 + az + bz1 + z2 = –a and z1 z2 = bAgain 0, z1, z2 are vertices of an equilateral triangle.Therefore, 02 + z12 + z22 = 0z1 + z1z2 + 0z2 = 0z12 + z22 = z1z2(z1 + z2)2 = 3z1z2
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached solution to your problem below.

Given that z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex.
So, we have z1 + z2 = -a, z1z2 = b (sum and product of roots)
So, z2 = z1 ( cos 60° + i sin 60°) = z1 (1/2 + √3/2 i)
So, 2z2 – z1 = √3 i z1
This gives, (2z2 -z1)2 = -3z12
Hence, (z12 + z22) = z1z2
So, a2 – 2b = b
this gives a2 = 3b

Hope it helps.
Thanks and regards,
Kushagra

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