Question

if x2 + kx + 9 = 0 are real and equal then what is the value of k

Answer 1

$\large\bold{\underline{\underline{Correct \: Question:-}}}$

if x2 - kx + 9 = 0 are real and equal then what is the value of k

$\large\bold{\underline{\underline{To \: Find:-}}}$

The Value of K.

$\large\bold{\underline{\underline{Answer:-}}}$

The given quadric equation is x2 - kx + 9 = 0, and roots are real and distinct

Then find the value of k.

Here,

a = 1, b = (-k) and c = 9

As we know that D = b2 - 4ac

Putting the value of a = 1, b = (-k) and c = 9

D = (-k)2 - 4 x (1) x (9)

= k2 - 36

The given equation will have real and distinct roots, if D > 0

k2 - 36 > 0

Now factorizing of the above equation

k2 - 36 > 0

k2 > 36

k>36=±6

k < -6 Or k > 6

Therefore, the value of k < -6 Or k > 6.

Answer 2

Step-by-step explanation:

Rhizobia are diazotrophic bacteria that fix nitrogen after becoming established inside the root nodules of legumes. To express genes for nitrogen fixation, rhizobia require a plant host; they cannot independently fix nitrogen. In general, they are gram negative, motile, non-sporulating rods.