Question 997509
find the values of p for which {{{(7p+1)x^2+(5p-1)x+p=1}}} has equal roots

For the quadratic equation to have equal roots, the argument to the square root must equal zero and the solution is {{{-b / 2a}}}.

 {{{(7p+1)x^2+(5p-1)x+p=1}}}
 {{{(7p+1)x^2+(5p-1)x+p-1=0}}}

{{{b^2 - 4ac = 0 }}}
{{{a=(7p+1)}}}
{{{b=(5p-1)}}}
{{{c=p-1}}}

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

{{{(5p-1)^2 -4(7p+1)(p-1) = 0}}}

{{{25p^2 -10p+1-(28p^2 +4)(p-1) = 0}}}

{{{25p^2 -10p+1-(28p^3-28p^2+4p-4) = 0}}}

{{{25p^2 -10p+1-28p^3+28p^2-4p+4 = 0}}}

 {{{-3p^2+14p+5 = 0}}}...........using quadratic formula, you have

{{{p= (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{p= (-14 +- sqrt( 14^2-4*(-3)*5 ))/(2*(-3)) }}}

{{{p= (-14 +- sqrt( 196+60 ))/-6 }}}

{{{p= (-14 +- sqrt( 256 ))/-6 }}} 

{{{p= (-14 +- 16)/-6 }}}

{{{p= (-cross(14)7 +- cross(16)8)/cross(-6)-33 }}} 

{{{p= (-7 +- 8)/-3 }}}    

solutions:

{{{p= (-7 + 8)/-3 }}} =>{{{p= -1/3 }}}
or
{{{p= (-7 - 8)/-3 }}}=>{{{p= -15/-3 }}}=>{{{p= 5}}}


then your equation will be

if{{{ p =-1/3}}}

{{{(7(-1/3)+1)x^2+(5(-1/3)-1)x+(-1/3)-1=0}}}

{{{(-7/3+3/3)x^2+(-5/3-3/3)x-1/3-3/3=0}}}

{{{-(4/3)x^2-(8/3)x-4/3=0}}} => this works, gives us one  root

check:

{{{ graph( 600, 600, -10, 10, -10, 10, -(4/3)x^2-(8/3)x-4/3) }}} 


or

if {{{p =5}}}

{{{(7(5)+1)x^2+(5(5)-1)x+5-1=0}}}

{{{36x^2+24x+4=0}}}=> this works, gives us one  root



check:

{{{ graph( 600, 600, -5, 5, -10, 10, 36x^2+24x+4) }}}