Question 1179144


 For what values of {{{c }}}will {{{x^2 +28x + c = 0}}} have {{{no}}}{{{ real}}} solutions? Explain

{{{x^2 +28x + c = 0}}}

use discriminant {{{b^2-4ac}}} to find {{{c}}}


recall, if {{{b^2-4ac<0}}},   there will be {{{no }}}{{{real}}} solutions

in your case {{{a=1}}}, {{{b=28}}}, and {{{c=c}}}

then

{{{28^2-4*1*c<0}}}

{{{28^2< 4c}}}

{{{784< 4c}}}

{{{784/4< c}}}

{{{196< c}}}

or

{{{c >196}}}



check: let {{{c=197}}}

{{{x^2 +28x + 197= 0}}}

using calculator we get

Complex solutions:
{{{x = -14 - i}}}
{{{x = -14 +i}}}