Question 928877
I guess you have quadratic equation:
 
{{{(k+4)x^2+(k+1)x+1=0}}} 

in the quadratic equation {{{ax^2 +bx +c=0}}} the roots are equal when
the discriminant {{{b^2-4ac = 0}}}

comparing the equation
{{{a= (k+4)}}}
{{{ b= (k+1)}}}
{{{ c=1}}}

so,

{{{(k+1)^2-4(k+4)(1)=0}}}

{{{k^2+2k+1-4k-16=0}}}

{{{k^2-2k-15=0}}}

{{{k^2-5k+3k-15=0}}}

{{{k(k-5)+3(k-5)=0}}}

{{{(k-5)(k+3)=0}}}

{{{k=5}}} or {{{k=-3 }}}


check the roots:

{{{(k+4)x^2+(k+1)x+1=0}}} if {{{k=5}}}

{{{(5+4)x^2+(5+1)x+1=0}}}

{{{9x^2+6x+1=0}}}

{{{9x^2+3x+3x+1=0}}}

{{{(9x^2+3x)+(3x+1)=0}}}

{{{3x(3x+1)+(3x+1)=0}}}

{{{(3x+1)(3x+1)=0}}}...roots will be same


{{{3x+1=0}}}=>{{{x=-1/3}}}... double solution


{{{ graph( 600, 600, -5, 5, -5, 10, (3x+1)(3x+1)) }}}