Question 1184307


Given that the equation {{{px^2 + 3px + p + q = 0}}},where {{{p <> 0}}}, has two equal real roots, find {{{q}}} in terms of {{{p}}}.


 the equation has two equal real roots if discriminant is equal to zero

{{{b^2-4ac=0}}}

{{{px^2 + 3px + p + q = 0}}} in your case {{{a=p}}}, {{{b=3p}}}, and {{{c= p + q }}}

{{{(3p)^2-4*p*(p + q)=0}}}

{{{9p^2- (4p^2 + 4pq)=0}}}

{{{9p^2- 4pq- 4p^2=0}}}

{{{5p^2- 4pq=0}}}...factor

{{{p (5p - 4q) = 0}}}


solutions:

{{{p=0}}}-> exclude because gien that {{{p <> 0}}}

{{{(5p - 4q) = 0}}}=>{{{5p = 4q}}}=>{{{5p =4q}}}=>{{{q=(5/4)p  }}}


{{{px^2 + 3px + p + q = 0}}}.......substitute {{{q=(5/4)p }}}

{{{px^2 + 3px + p + (5/4)p = 0}}}.....both sides multiply by {{{4}}}

{{{4px^2 + 12px + 4p + 5p = 0}}}

{{{4px^2 + 12px + 9p = 0}}}

check if discriminant is equal to zero:

{{{(12p)^2-4(4p)(9p)=0}}}

{{{144p^2-144p^2=0}}}

{{{0=0}}}=>true