Question 1184293


Is this correct statement that equations 
{{{x^2+x+a=0 }}}and 
{{{x^2+ax +1=0 }}}
have a common real point for exactly one value of {{{a}}}"?

check  if there is a value of {{{a}}} for what discriminants are equal to zero

{{{x^2+x+a=0}}}

 discriminant: 

{{{b^2-4ac=0}}}.........in your case {{{a=1}}}, {{{b=1}}} and {{{c=a}}}
{{{1^2-4*1*a=0}}}
{{{1 -4a=0}}}
{{{1 =4a }}}
{{{a=1/4}}}


{{{x^2+ax +1=0}}}

 discriminant: 

{{{b^2-4ac=0}}}.........in your case{{{ a=1}}}, {{{b=a}}} and {{{c=1}}}
{{{a^2-4*1*1=0}}}
{{{a^2-4=0}}}
{{{a^2=4}}} 
{{{a=2}}} or{{{ a=-2}}}


check if any of {{{a}}} will give us one solution:

{{{a=1/4}}}
{{{x^2+x+1/4=0 }}} 
{{{x^2+(1/4)x +1=0}}}
________________________-> no solution exist

{{{a=2}}}
{{{x^2+x+2=0 }}} 
{{{x^2+2x +1=0}}}
_______________-> no solution exist

{{{a=-2}}}
{{{x^2+x-2=0 }}} 
{{{x^2-2x +1=0}}}
_______________ ->solution {{{exist}}}

parabolas have a common real point for exactly one value of {{{a=-2}}}

so, the statement above is correct