SOLUTION: For what values of c will x² +28x + c = 0 have no real solutions? Explain

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Question 1179144: For what values of c will x² +28x + c = 0 have no real solutions? Explain
Found 3 solutions by mananth, MathLover1, ikleyn:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
x² +28x + c = 0
comparing with ax2+bx+c=1


a=1, b=28 , c=1
b^2 -4ac = 28^2 - 4*1*1
784/4 = 196
c>196
For no real solutions b^2-4ac <0

Answer by MathLover1(20854) About Me  (Show Source):
You can put this solution on YOUR website!

For what values of c+will x%5E2+%2B28x+%2B+c+=+0 have no+real solutions? Explain
x%5E2+%2B28x+%2B+c+=+0
use discriminant b%5E2-4ac to find c

recall, if b%5E2-4ac%3C0, there will be no+real solutions
in your case a=1, b=28, and c=c
then
28%5E2-4%2A1%2Ac%3C0
28%5E2%3C+4c
784%3C+4c
784%2F4%3C+c
196%3C+c
or
c+%3E196


check: let c=197
x%5E2+%2B28x+%2B+197=+0
using calculator we get
Complex solutions:
x+=+-14+-+i
x+=+-14+%2Bi


Answer by ikleyn(53618) About Me  (Show Source):
You can put this solution on YOUR website!
.
For what values of c will x^2 +28x + c = 0 have no real solutions? Explain
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The given quadratic equation has no real solutions if and only if its discriminant is a negative value.


The discriminant is  d = 28^2 - 4c = 784 - 4c.


The condition of negativity the discriminant is

    784 - 4c < 0,   or   4c > 784,   c > 784/4 = 196.


ANSWER.  The given equation has no real solutions at c > 196.

Solved.