Question 4919
These radical equations you solve by using the following property of equality:
For all real numbers a and b, if a=b, then {{{a^2=b^2}}}
A)
{{{sqrt(x+1)=5-sqrt(x-4)}}} Now square both sides
{{{sqrt(x+1)^2=(5-sqrt(x-4))^2}}}
Let's look at each side seperately since it becomes kind of messy.
{{{sqrt(x+1)^2}}} = {{{x+1}}}
For the right side:
{{{(5-sqrt(x-4))(5-sqrt(x-4))}}} = {{{25-5sqrt(x-4)-5sqrt(x-4)+(x-4)}}} =
{{{25-10sqrt(x-4)+x+4}}}={{{x+21-10sqrt(x-4)}}}
Thus we get:
{{{x+1=x+21-10sqrt(x-4)}}} 
Add x to and substract 21 from both sides
{{{20=-10sqrt(x-4)}}}
Divide both sides by -10
{{{-2=sqrt(x-4)}}} Again square both sides
{{{4=x-4}}} Add 4 to both sides
{{{x=8}}}
For the next 2 you proceed the same pattern