Question

The smallest value of k for which the equation x square + kx + 9 equal to zero has real root is

Answer 1

Answer:

since the eqn has real roots

∴ b²-4ac=0

⇒k²-4×1×9=0

⇒k²=√36

⇒k=±6

∵the smallest value of k is required

∴k= -6

thank u........................

Answer 2

k ≥ ± 6

Step-by-step explanation:

The given quadratic equation:

$x^{2} +kx+9=0$

To find, the value of k = ?

Here, a = 1, b = k and c = 9

∴ $D=b^{2}-4ac$

$=k^{2}-4(1)(9)$

$=k^{2}-36$

The roots are real, then D ≥ 0.

∴$k^{2}-36$ ≥ 0

⇒ [tex]k^{2}-36 ≥ 0

⇒[tex]k^{2} ≥3

⇒ k ≥ ± 6

Hence,  k ≥ ± 6