JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 19)
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 :
Answer
6
Explanation
For equilateral triangle with vertices z1, z2 and z3,
$$z_1^2 + z_2^2 + z_3^3 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}$$
Here one vertex z3 is 0
$$ \therefore $$ $$z_1^2 + z_2^2 = {z_1}{z_2} + 0 + 0$$
Given, z1, z2 are roots of $${z^2} + az + 12 = 0$$
$$ \therefore $$ $${z_1} + {z_2} = - a$$
$${z_1}{z_2} = 12$$
$$ \therefore $$ $$z_1^2 + z_2^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2}$$
$$ \Rightarrow {({z_1} + {z_2})^2} = 3{z_1}{z_2}$$
$$ \Rightarrow {( - a)^2} = 3 \times 12$$
$$ \Rightarrow {a^2} = 36$$
$$ \Rightarrow a = \pm 6$$
$$ \Rightarrow |a|\, = 6$$
$$z_1^2 + z_2^2 + z_3^3 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}$$
Here one vertex z3 is 0
$$ \therefore $$ $$z_1^2 + z_2^2 = {z_1}{z_2} + 0 + 0$$
Given, z1, z2 are roots of $${z^2} + az + 12 = 0$$
$$ \therefore $$ $${z_1} + {z_2} = - a$$
$${z_1}{z_2} = 12$$
$$ \therefore $$ $$z_1^2 + z_2^2 + 2{z_1}{z_2} = {z_1}{z_2} + 2{z_1}{z_2}$$
$$ \Rightarrow {({z_1} + {z_2})^2} = 3{z_1}{z_2}$$
$$ \Rightarrow {( - a)^2} = 3 \times 12$$
$$ \Rightarrow {a^2} = 36$$
$$ \Rightarrow a = \pm 6$$
$$ \Rightarrow |a|\, = 6$$
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