Question

Find K such that x^2 +K+4=0 shall be:
1. Real and Equal
2. Real and Different
3.Complex

Answer 1

Real and Equal

Answer 2

use the discriminant ... do u know how?

Answer 3

REAL AND EQUAL D=0

Answer 4

I do, thank you.

Answer 5

okee dokee

Answer 6

I have just reached a point of uncertainty. Could you assist me a bit?

Answer 7

yes

Answer 8

Could you guide me through coming to my Real and equal answer?
I used the discriminant coming up with
\[(-2)^2-4(K)(3)\]
What am I missing here?

Answer 9

D=b^2-4ac

Answer 10

b is 0 - no x term

Answer 11

Well if this is correct, I have K=32. Somewhere around the ballpark?

Answer 12

D= 0 - 4(1)(k+4)=0 and solve for k

Answer 13

-4k -16 =0 , k =-4

Answer 14

so... Discriminant is the quantity under the √ in the quadratic formula - b^2-4ac, in your case a = 1 , b = 0 , c = K+4

Answer 15

real and equal D=0, real and different D>0, complex D<0

Answer 16

How would I know that I need to put it in that format though?

Answer 17

the discriminant determines the nature of the roots

Answer 18

[i.e. D=0-4(1)(k+4)]

I'm not really sure how you came about it really. The way my professor taught it, he just gave us the natures and what you just said: real and equal D=0, real and different D>0, complex D<0

Answer 19

so what is D?

Answer 20

x^2 +0x+(K+4)=0 ---- this is the complete equation

Answer 21

D is the discriminant?

Answer 22

yes

Answer 23

which is equal to b^2-4ac ... but your b is 0

Answer 24

do you understand ?

Answer 25

Yes I am starting to.
So in 0-4(1)(k+4)=0
What made you use k+4 in a parenthesis?

Answer 26

so I can make it obvious that K+4 is c

Answer 27

did you understand now?