Question 1179332


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

a. two real number solutions

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=15}}}

{{{4x^2+16x+15}}}

and real solution is

{{{4x^2+16x+15=0}}} use quadratic formula

{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}............{{{a=4}}}, {{{b=16}}}, and {{{c=15}}}

{{{x=(-16+-sqrt(16^2-4*4*15))/(2*4)}}}

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

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

{{{x=(-4+-1)/2}}}

{{{x=-3/2}}}
or
{{{x=-5/2}}}

see the graph below: 


{{{ graph( 600, 600, -10, 10, -10, 10,4x^2+16x+15) }}}