SOLUTION: Please, please help me solve this equation, I can't seem to figure it out: {{{sqrt(t+3)-sqrt(t-2)=sqrt(7-t)}}}

Algebra ->  Radicals -> SOLUTION: Please, please help me solve this equation, I can't seem to figure it out: {{{sqrt(t+3)-sqrt(t-2)=sqrt(7-t)}}}      Log On


   

Question 57615This question is from textbook Elementary & Intermediate Algebra
: Please, please help me solve this equation, I can't seem to figure it out:
sqrt%28t%2B3%29-sqrt%28t-2%29=sqrt%287-t%29 This question is from textbook Elementary & Intermediate Algebra

Found 2 solutions by faceoff57, funmath:
Answer by faceoff57(108) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt(t+3)-sqrt(t-2)=sqrt(7-t)
{sqrt(t+3)-sqrt(t-2)=sqrt(7-t)}^2
(t+3)-(t-2)=7-t
t+3-t+2=7-t
5=7-t
t=7-5
t=2

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
I don't know who the other tutor was that answered your question, but they made a grave common error. They were right in that you have to square both sides, but he or she didn't square the left side properly. Here's the way you really solve it:
sqrt%28t%2B3%29-sqrt%28t-2%29=sqrt%287-t%29
%28sqrt%28t%2B3%29-sqrt%28t-2%29%29%5E2=%28sqrt%287-t%29%29%5E2 You have to FOIL the left.
%28sqrt%28t%2B3%29-sqrt%28t-2%29%29%28sqrt%28t%2B3%29-sqrt%28t-2%29%29=%287-t%29

%28t%2B3%29-2sqrt%28t%2B3%29%28t-2%29%2B%28t-2%29=7-t
t%2Bt%2B3-2-2sqrt%28%28t%2B3%29%28t-2%29%29=7-t
2t%2B1-2sqrt%28%28t%2B3%29%28t-2%29%29=7-t
-2t%2B2t%2B1-1-2sqrt%28%28t%2B3%29%28t-2%29%29=7-1-t-2t
-2sqrt%28%28t%2B3%29%28t-2%29%29=6-3t Square both sides again:
%28-2sqrt%28%28t%2B3%29%28t-2%29%29%29%5E2=%286-3t%29%5E2
4%28t%2B3%29%28t-2%29=%286-3t%29%286-3t%29
4%28t%5E2-2t%2B3t-6%29=36-18t-18t%2B9t%5E2
4%28t%5E2%2Bt-6%29=36-36t%2B9t%5E2
4t%5E2%2B4t-24=36-36t%2B9t%5E2
4t%5E2-4t%5E2%2B4t-4t-24%2B24=36%2B24-36t-4t%2B9t%5E2-4t%5E2
0=60-40t%2B5t%5E2 Write in a form you're used to factoring.
5t%5E2-40t%2B60=0 Factor out the GCF, 5:
5%28t%5E2-8t%2B12%29=0 Factor what's left in the parenthesis.
5%28t-6%29%28t-2%29=0
%285%2F5%29%28t-6%29%28t-2%29=0%2F5
%28t-6%29%28t-2%29=0 Set each parenthesis = to 0 and solve for t.
t-6=0 and t-2=0
t=6 and t=2
With these kinds of problems, you have to check for extraneous solutions by substituting t=6 and t=2 into the original equation and see if you're right.
sqrt%286%2B3%29-sqrt%286-2%29=sqrt%287-6%29
sqrt%289%29-sqrt%284%29=sqrt%281%29
3-2=1
1=1 t=6 checks out.
:
sqrt%282%2B3%29-sqrt%282-2%29=sqrt%287-2%29
sqrt%285%29-sqrt%280%29=sqrt%285%29
sqrt%285%29=sqrt%285%29 t=2 checks out also, so the solutions really are:
highlight%28t=6%29 and highlight%28t=2%29
:
I hope that last would be tutor didn't mess up your thinking. You CAN'T do what he or she did. It's a common mistake that you may not even get partial credit for.
Happy Calculating!!!