The origin and roots of the equation #z^2 +pz + q=0# form an equilateral triangle. What is the relation between p and q?
2 Answers
Explanation:
The roots are of the form
So we have:
#z^2+pz+q = (z-ksqrt(3)/2-k1/2i)(z-ksqrt(3)/2+k1/2i)#
#color(white)(z^2+pz+q) = z^2-ksqrt(3)z+k^2#
So:
#p^2 = 3q#
Explanation:
Suppose we allow any value for
The two roots can generally be expressed as:
#alpha = r (cos(theta + pi/6) + i sin(theta + pi/6))#
#beta = r (cos(theta - pi/6) + i sin(theta - pi/6))#
Note that:
#cos A + cos B = 2 cos((A+B)/2)cos((A-B)/2)#
#sin A + sin B = 2 sin((A+B)/2)cos((A-B)/2)#
Then:
#alpha+beta = r(cos(theta + pi/6) + cos(theta - pi/6)) + i r(sin(theta + pi/6) + sin(theta-pi/6))#
#color(white)(alpha+beta) = 2r cos(theta)cos(pi/6) + 2i r sin(theta )cos(pi/6)#
#color(white)(alpha+beta) = sqrt(3) r(cos(theta)+i sin(theta))#
#alpha beta = r^2(cos(theta + pi/6) + i sin(theta + pi/6))(cos(theta - pi/6) + i sin(theta - pi/6))#
#color(white)(alpha beta) = r^2(cos(2theta) + i sin(2theta))#
#color(white)(alpha beta) = (r(cos(theta) + i sin(theta)))^2#
#color(white)(alpha beta) = 1/3(alpha + beta)^2#
#z^2+pz+q = (z-alpha)(z-beta)#
#color(white)(z^2+pz+q) = z^2-(alpha+beta)z+alphabeta#
So equating coefficients, we find:
#p^2 = (-(alpha+beta))^2 = 3(alpha beta) = 3q#