Question

For what value (s) of A quadratic equation 3x^2 -6x +1 = 0 has no real roots ?​

Answer 1

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Answer 2

Answer:

For what value (s) of A quadratic equation 3x^2 -6x +1 = 0 has no real roots ?

=> k ≤ -4 and k ≥ 4

Step-by-step explanation:

Discriminant of quadratic equation is equal to zero, or more than zero then roots are real.

3x² - k√3x + 4 = 0

Compare with ax + bx + c = 0, then a = 3, -k√3 and c = 4

D = b² - 4ac

For real roots b2 - 4ac ≥ 20

(-k√3)² - 4x3x4 ≥ 0

3k² - 48 ≥ 0

K² - 16 ≥ 0

(k- 4)(k + 4) ≥ 0

k ≤ -4 and k ≥ 4

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