JEE MainMathematicsModerateTheory of Equations

For the equation 3x^2 + px + 3 = 0, p > 0 if one of the roots
is square of the other, then p is equal to

Show answer and solution

3x^{2} + px + 3 = 0;p > 0

Let one root be

\alpha

.
Other root

= \alpha^{2}

\alpha \times \alpha^{2} = 3/3

(Product of roots)

\Rightarrow \alpha^{3} = 1

\Rightarrow \alpha = 1,

w, w^2 (cube root of unity)
Now,

\alpha + \alpha^{2} = P/3

(sum of roots)

\Rightarrow w + w^{2} = - P/3

\because

1 + w + w^2 = 0 (Property of cube root of unity)

\Rightarrow w + w^{2} = - 1

\Rightarrow - 1 = - P/3

\Rightarrow P = 3

Source: Grass-G / M-06 Mathematics Paper 14 October.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.