Question

For what value(s) of ‘a’ quadratic equation 30ax² − 6x + 1 = 0 has no real
roots?​

Answer 1

Step-by-step explanation:

for no real roots, D = b² - 4ac < 0

here a = 30a, b = -6 and c = 1

therefore D = (-6)² - ( 4×30a×1) < 0

=> 36 - 120a < 0

=> 36 < 120a

=> 120a > 36

=> a > 36/120

=> a > 3/10

therefore for all values of a greater than 3/10, the given quadratic equation will have imaginary roots