Question 159293
Note: repeated roots simply mean that the roots are equal. This only happens when the discriminant D is zero
 


Looking at {{{(p-1)x^2 + 6x + (p+7) = 0 }}}, we see that {{{a=p-1}}}, {{{b=6}}} and {{{c=p+7}}}



{{{D=b^2-4ac}}} Start with the discriminant formula



{{{0=6^2-4(p-1)(p+7)}}} Plug in {{{D=0}}} (since the roots repeat themselves), {{{a=p-1}}}, {{{b=6}}} and {{{c=p+7}}}



{{{0=36-4(p-1)(p+7)}}} Square 6 to get 36



{{{0=36-4(p^2+6p-7)}}} FOIL



{{{0=36-4p^2-24p+28}}} Distribute



{{{0=-4p^2-24p+64}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ap^2+bp+c}}} where {{{a=-4}}}, {{{b=-24}}}, and {{{c=64}}}



Let's use the quadratic formula to solve for p



{{{p = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{p = (-(-24) +- sqrt( (-24)^2-4(-4)(64) ))/(2(-4))}}} Plug in  {{{a=-4}}}, {{{b=-24}}}, and {{{c=64}}}



{{{p = (24 +- sqrt( (-24)^2-4(-4)(64) ))/(2(-4))}}} Negate {{{-24}}} to get {{{24}}}. 



{{{p = (24 +- sqrt( 576-4(-4)(64) ))/(2(-4))}}} Square {{{-24}}} to get {{{576}}}. 



{{{p = (24 +- sqrt( 576--1024 ))/(2(-4))}}} Multiply {{{4(-4)(64)}}} to get {{{-1024}}}



{{{p = (24 +- sqrt( 576+1024 ))/(2(-4))}}} Rewrite {{{sqrt(576--1024)}}} as {{{sqrt(576+1024)}}}



{{{p = (24 +- sqrt( 1600 ))/(2(-4))}}} Add {{{576}}} to {{{1024}}} to get {{{1600}}}



{{{p = (24 +- sqrt( 1600 ))/(-8)}}} Multiply {{{2}}} and {{{-4}}} to get {{{-8}}}. 



{{{p = (24 +- 40)/(-8)}}} Take the square root of {{{1600}}} to get {{{40}}}. 



{{{p = (24 + 40)/(-8)}}} or {{{p = (24 - 40)/(-8)}}} Break up the expression. 



{{{p = (64)/(-8)}}} or {{{p =  (-16)/(-8)}}} Combine like terms. 



{{{p = -8}}} or {{{p = 2}}} Simplify. 



So the answers are {{{p = -8}}} or {{{p = 2}}} 



This means that if {{{p = -8}}} or {{{p = 2}}}, then the equation {{{(p-1)x^2 + 6x + (p+7) = 0 }}} will have repeated roots.