Question

For what values of k, the equation kx²+6x+1=0 has real roots?​

Answer 1

9 > k

Step-by-step explanation:

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Answer 2

Answer:

For k < 9, the equation kx² + 6x + 1 = 0 has real roots.

Step-by-step explanation:

Every quadratic equation is in the form of

$a {x}^{2} + bx + c = 0$

where a ≠ 0.

Here we have

a = k

b = 6

c = 1

We know that Discriminant for real roots is

D = b² - 4ac > 0

by putting the values of a, b and c, we have

6² - 4k > 0

or 36 - 4k > 0

or 36 > 4k or 4k < 36

this implies,

$k < \frac{36}{4} \: or \: k < 9$

Therefore, for k < 9, the equation kx² + 6x + 1 = 0 has real roots.