Question 1179334


the values of {{{k}}} such that {{{4x^2+16x+k}}} has the following solutions.

a. no real  solution

it will be if discriminant {{{b^2-4ac<0}}}

{{{4x^2+16x+k}}}...in this case {{{a=4}}}, {{{b=16}}}, and {{{c=k}}}

{{{16^2-4*4*k<0}}}

{{{16^2-16k<0}}}

{{{16(16-k)<0}}}

will be equal to zero if {{{16-k<0}}}->{{{16<k}}} or {{{k>16}}}

let try {{{k=17}}}

{{{4x^2+16x+17}}}

and real solution is

{{{4x^2+16x+17=0}}} use quadratic formula

{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}............{{{a=4}}}, {{{b=16}}}, and {{{c=17}}}

{{{x=(-16+-sqrt(16^2-4*4*17))/(2*4)}}}

{{{x=(-16+-sqrt(-16))/8}}}

{{{x=(-16+-4i)/8}}}

{{{x=(-4+-i)/2}}}

{{{x=-2+i/2}}}
or
{{{x=-2-i/2}}}

see the graph below: 


{{{ graph( 600, 600, -10, 10, -10, 10,4x^2+16x+17) }}}