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
Jan 21, 2018

#p^2=3q#

Explanation:

The roots are of the form #k(sqrt(3)/2+-1/2i)# for some constant #k#. If #p, q# are real then #k# is real. Otherwise it can be any complex constant.

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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#

Jan 21, 2018

#p^2=3q#

Explanation:

Suppose we allow any value for #p# and #q# including Complex values.

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#