Q4 of 227 Page 110

Show that the roots of the equation 3p2x2 – 2pqx + q2 = 0 are not real.

3p2x2 – 2pqx + q2 = 0

Compare this equation with ax2 + bx + c = 0


∴ a = 3p2 , b = 2pq and c = q2


b2 – 4ac = (2pq)2 – 4(3p2)[q2]


= 4p2q2 – 12p2q2


= – 8p2q2


Since squared quantity is always positive.


Hence, p2q2 ≥ 0


Now, it is given p ≠ q, so p2q2 > 0


So, D = –8p2q2 will be negative.


Hence the equation has no real roots.


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