Question

For what value of ‘a’ quadratic equation ax

2

 6x + 1 = 0 has no real roots ?​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

3ax2 - 6x + 1 = 0  (- 6)2 - 4(3a) (1) < 0  12a > 36 => a > 3

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