Question 1177572
{{{q}}} is missing from given equation, so I assume you have{{{ x^2+ px +4q = 0}}}


a) Find the values of {{{p}}} and {{{q}}}.


use the rule: sum of the roots = {{{-b/a }}}
roots are {{{-1}}} and {{{4}}}, sum is {{{-1+4=3}}}

{{{-b/a =3}}}.........{{{b=p}}}, {{{a=1}}}
{{{-p/1 =3}}}
{{{p=-3}}}

and 

product of the roots ={{{ c/a}}}



{{{c=4q}}}, {{{a=1}}}

{{{4q/1=(-1)*4}}}

{{{4q=-4}}}

{{{q=-1}}}

{{{ x^2-3x -4 = 0}}}


b) Using these values of {{{p }}}and {{{q}}} find the value of {{{r}}}, where {{{r}}} is a constant, and the equation 
{{{x^2+ px + r=0}}} has {{{equal}}} roots 

{{{p=-3}}}

{{{x^2-3x + r=0}}}.....if equation has {{{equal}}} roots, write left side as square of difference
sum of the roots {{{-b/a =-(-3)/1 =3}}}

{{{x[1]+x[2]=3}}}..............eq.1. since {{{x^2+ px + r=0}}} has {{{equal}}} roots, {{{x[1]=x[2]}}}

{{{2x[1]=3}}}
{{{x[1]=3/2}}}

product of the roots ={{{ c/a=r/1=r}}} 

{{{x[1]*x[1]=r}}}..............eq.2
{{{(3/2)*(3/2)=r}}}
{{{r=9/4}}}

and your equatio is:

{{{x^2-3x + 9/4=0}}}