Question 56523
{{{sqrt(4p+5) + sqrt (p+5) = 3}}}
{{{sqrt(4p+5)+sqrt(p+5)-sqrt(p+5)=3-sqrt(p+5)}}}
{{{sqrt(4p+5)=3-sqrt(p+5)}}}
{{{(sqrt(4p+5))^2=(3-sqrt(p+5))^2}}}  The rule is {{{highlight((sqrt(a))^2=a)}}}
{{{4p+5=(3-sqrt(p+5))(3-sqrt(p+5))}}} FOIL
{{{4p+5=3(3)+3(-sqrt(p+5))-sqrt(p+5)(3)-sqrt(p+5)(-sqrt(p+5))}}}
{{{4p+5=9-3sqrt(p+5)-3sqrt(p+5)+(p+5)}}}
{{{4p+5=-6sqrt(p+5)+p+14}}}
{{{4p-p+5-14=-6sqrt(p+5)+p-p+14-14}}}
{{{3p-9=-6sqrt(p+5)}}}
{{{3(p-3)=-6sqrt(p+5)}}}
{{{(3/3)(p-3)=(-6/3)sqrt(p+5)}}}
{{{(p-3)=-2*sqrt(p+5)}}}
{{{((p-3))^2=(-2*sqrt(p+5))^2}}}
{{{p^2-6p+9=4(p+5)}}}  This is a quadratic equation and needs to be set =0 and factored if possible.
{{{p^2-6p+9=4p+20}}}
{{{p^2-6p-4p+9-20=4p-4p+20-20}}}
{{{p^2-10p-11=0}}}
(p________)(p______)=0
We need to fill the blanks with two integers that multiply to get -11, but add to get -10:
(-11)*(+1)=-11
(-11)+(+1)=-10
 (p-11)(p+1)=0  Set each parenthesis =0 and solve for x:
p-11=0  
p=11
P+1=0
p=-1
So x=11 and x=-1 are possible solutions.  However, with even radical equations , rational equations, and log equations we have to check our answers for false solutions called  extraneous solutions.
substituting x=11
{{{sqrt(4*highlight(11)+5)+sqrt(highlight(11)+5)=3}}}
{{{sqrt(44+5)+sqrt(11+5)=3}}}
{{{sqrt(49)+sqrt(16)=3}}}
{{{7+4=3}}}  
11=3  This proves that x=11 is an extraneous solution.
 Substituting x=-1
{{{sqrt(4*highlight(-1)+5)+sqrt(highlight(-1)+5)=3}}}
{{{sqrt(-4+5)+sqrt(-1+5)=3}}}
{{{sqrt(1)+sqrt(4)=3}}}
{{{1+2=3}}}
3=3  This proves that x=-1 is a solution to the equation:
{{{highlight(x=-1)}}}
Happy Calculating!!!