Math, asked by dharan1803, 8 months ago

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

Attachments:

Answers

Answered by karankirat345
9

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!!!

Answered by dharaneeshkumar1803
1

Answer:

1) - root 16

2) + or - 1

hope it helps

Similar questions