SOLUTION: I need some help please: Solve the following equations: A) sqr(x+1)= 5- sqr(x-4) and B) sqr(x-2)=sqr(x+7)-1 and C) sqr(2n-1) + sqr(n-5) = 3 Please with explanation. Thanks

Algebra ->  Algebra  -> Square-cubic-other-roots -> SOLUTION: I need some help please: Solve the following equations: A) sqr(x+1)= 5- sqr(x-4) and B) sqr(x-2)=sqr(x+7)-1 and C) sqr(2n-1) + sqr(n-5) = 3 Please with explanation. Thanks      Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 4919: I need some help please: Solve the following equations: A) sqr(x+1)= 5- sqr(x-4) and B) sqr(x-2)=sqr(x+7)-1 and C) sqr(2n-1) + sqr(n-5) = 3
Please with explanation. Thanks
Answer by guapa(62) About Me  (Show Source):
You can put this solution on YOUR website!
These radical equations you solve by using the following property of equality:
For all real numbers a and b, if a=b, then a%5E2=b%5E2
A)
sqrt%28x%2B1%29=5-sqrt%28x-4%29 Now square both sides
sqrt%28x%2B1%29%5E2=%285-sqrt%28x-4%29%29%5E2
Let's look at each side seperately since it becomes kind of messy.
sqrt%28x%2B1%29%5E2 = x%2B1
For the right side:
%285-sqrt%28x-4%29%29%285-sqrt%28x-4%29%29 = 25-5sqrt%28x-4%29-5sqrt%28x-4%29%2B%28x-4%29 =
25-10sqrt%28x-4%29%2Bx%2B4=x%2B21-10sqrt%28x-4%29
Thus we get:
x%2B1=x%2B21-10sqrt%28x-4%29
Add x to and substract 21 from both sides
20=-10sqrt%28x-4%29
Divide both sides by -10
-2=sqrt%28x-4%29 Again square both sides
4=x-4 Add 4 to both sides
x=8
For the next 2 you proceed the same pattern