z2+az+12=0
z1+z2=−a and z1z2=12
If 0,z1,z2 are vertices of equilateral triangles
z2=z1e3iπ
z2=z1(21+i23)
2z2−z1=3iz1
squaring both sides
⇒4z22+z12−4z1z2=−3z12
⇒(z1+z2)2=3z1z2
⇒a2=3×12
⇒∣a∣=6
Let z1,z2 be the roots of the equation z2+az+12=0 and z1,z2 form an equilateral triangle with origin. Then, the value of ∣a∣ is
Held on 18 Mar 2021 · Verified 6 Jul 2026.
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