Question

values of k for which the equation 9x2+2kx+1=0 has real roots

Answer 1

it have real roots
So,
$b ^{2} - 4ac > 0$
$(2k) ^{2} - 4 \times 1 \times 9 > 0$
4k^2 - 36 > 0 = 4k^2 >36
k^2 >18 = k > √18
hence k = √18
well not sure may be

Answer 2

Answer:

Step-by-step explanation:

D=b²-4ac

(2k)²- 4×-1×9

4k²+ 36

We know that D is greater than 0 as it has real roots.

So

4k²+36>0

k²+9>0