Question

pls solve this maths question
(9 or 10 or both) its upto ur ability​

Answer 1

Answer:

9.

$for \: real \: roots \\ d > 0 \\ {b}^{2} - 4ac > 0 \\ {k}^{2} - 4 \times 1 \times - 4 > 0 \\ {k}^{2} + 16 > 0 \\ {k}^{2} > - 16 \\ k > \sqrt{ - 16}$

10.

$for \: real \: roots \\ d > 0 \\ {b}^{2} - 4ac > 0 \\ { - 2}^{2} - 4 \times k \times k > 0 \\ 4 - 4 {k}^{2} > 0 \\ - 4 {k}^{2} > - 4 \\ {k}^{2} > \frac{ - 4}{ - 4} \\ {k}^{2} > 1 \\ k > \sqrt{1} \\ k > + 1 \: or - 1$

Hope this helps you...

Be Brainly!!!

Answer 2

Answer:

1) - root 16

2) + or - 1

hope it helps