SOLUTION: Hello, I know the answer is 14 to the following problem but I can't figure out how to solve it: {{{ 2sqrt(x+11)=sqrt(x+2)+sqrt(3x-6) }}} I first tried to square both sides: {

Algebra ->  Square-cubic-other-roots -> SOLUTION: Hello, I know the answer is 14 to the following problem but I can't figure out how to solve it: {{{ 2sqrt(x+11)=sqrt(x+2)+sqrt(3x-6) }}} I first tried to square both sides: {      Log On


   

Question 856704: Hello,
I know the answer is 14 to the following problem but I can't figure out how to solve it:
+2sqrt%28x%2B11%29=sqrt%28x%2B2%29%2Bsqrt%283x-6%29+
I first tried to square both sides:
+%284%29%28x%2B11%29+=+x%2B2%2B2sqrt%28x%2B2%29sqrt%283x-6%29%2B3x-6+
simplify:
+4x%2B44+=+4x+-+4+%2B2sqrt%28x%2B2%29sqrt%283x-6%29+ --->
+48+=+2sqrt%28x%2B2%29sqrt%283x-6%29+ --->
+24+=+sqrt%28x%2B2%29sqrt%283x-6%29+
Now what??? square both sides again?
Thanks.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You are correct. You would square both sides to eliminate the square root. Here is one way to do it.


24+=+sqrt%28x%2B2%29sqrt%283x-6%29


24+=+sqrt%28%28x%2B2%29%283x-6%29%29 Look at Rule 1 under the "distributing" section from this page


24%5E2+=+%28sqrt%28%28x%2B2%29%283x-6%29%29%29%5E2 Square both sides


576+=+%28x%2B2%29%283x-6%29


576+=+x%283x-6%29%2B2%283x-6%29


576+=+3x%5E2-6x%2B6x-12


576+=+3x%5E2-12


3x%5E2-12+=+576


3x%5E2+=+576%2B12


3x%5E2+=+588


x%5E2+=+588%2F3


x%5E2+=+196


x+=+%22%22%2B-sqrt%28196%29


x+=+sqrt%28196%29 or x+=+-sqrt%28196%29


x+=+14 or x+=+-14


So the possible solutions are x+=+14 or x+=+-14


I'll leave the check to you to do. You need to plug each possible answer into the original equation and simplify. If you get a true statement, then the possible solution is a real solution. If not, then it's not a real solution.