You visited us 1 times! Enjoying our articles? Unlock Full Access!

Orthocentre

The origin an...

Question

The origin and the roots of the equation $z^{2} + p z + q = 0$ form an equilateral triangle if

p^{2} = 2 q

No worries! We‘ve got your back. Try BYJU‘S free classes today!

p^{2} = q

No worries! We‘ve got your back. Try BYJU‘S free classes today!

p^{2} = 3 q

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

q^{2} = 3 p

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $p^{2} = 3 q$
$z^{2} + p z + q = 0$
$z_{1} + z_{2} = p, z_{1} z_{2} = q$
If $z_{1}, z_{2}, z_{3}$ are the vertices of an equilateral triangle then $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}$
If $z_{3}$ is origin then $z_{1}^{2} + z_{2}^{2} = z_{1} z_{2}$
$\Rightarrow (z_{1} + z_{2})^{2} = 3 z_{1} z_{2}$
$\Rightarrow p^{2} = 3 q$

Suggest Corrections

Similar questions

z_{1}

z_{2}

z^{2} + p z + q = 0

p

q

z_{1}, z_{2}

Z_{1}, Z_{2}, Z_{3}

x^{3} + 3 p x^{2} + 3 q x + r = 0

z^{2} + p z + q = 0

s e c α, c o s e c α

x^{2} - p x + q = 0

p^{2} - 2 q - q^{2} = 0

x^{2} - x s i n α + c o s α = 0

p^{2} = \frac{1 - q^{2}}{1 + q^{2}}

z^{2} + 2 z + q = 0

q

Join BYJU'S Learning Program