SOLUTION: I have tried this problem a million times and I can not get the answer that the book gives, can someone please show me??? {{{sqrt(a+7)=1+sqrt(2a)}}} the book says that the answer

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Question 60809: I have tried this problem a million times and I can not get the answer that the book gives, can someone please show me???
sqrt%28a%2B7%29=1%2Bsqrt%282a%29
the book says that the answer is 2
Found 2 solutions by funmath, hayek:
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28a%2B7%29=1%2Bsqrt%282a%29
%28sqrt%28a%2B7%29%29%5E2=%281%2Bsqrt%282a%29%29%5E2
a%2B7=%281%2Bsqrt%282a%29%29%281%2Bsqrt%282a%29%29
a%2B7=1%2B2sqrt%282a%29%2B2a
a-2a%2B7-1=1-1%2B2sqrt%282a%29%2B2a-2a
-a%2B6=2sqrt%282a%29
%28-a%2B6%29%5E2=%282sqrt%282a%29%29%5E2
%28-a%2B6%29%28-a%2B6%29=2%5E2%282a%29
a%5E2-12a%2B36=4%282a%29
a%5E2-12a%2B36=8a
a%5E2-12a-8a%2B36=8a-8a
a%5E2-20a%2B36=0
(a-2)(a-18)=0
a-2=0 and a-18=0
a=2 and a=18 Check for extraneous solutions:
sqrt%282%2B7%29=1%2Bsqrt%282%282%29%29
sqrt%289%29=1%2Bsqrt%284%29
3=1%2B2
3=3 a=2 is a valid solution.
sqrt%2818%2B7%29=1%2Bsqrt%282%2818%29%29
sqrt%2825%29=1%2Bsqrt%2836%29
5=1%2B6
5=7 a=18 is extraneous.
The only valid solution is a=2.
Happy Calculating!!!

Answer by hayek(51) About Me  (Show Source):
You can put this solution on YOUR website!
Problem: sqrt%28a%2B7%29=1%2Bsqrt%282a%29
One easy way to solve this is to square both sides of the equation. When you do this, however, you have to be careful to check the solutions you get, because squaring both sides will introduce potentially false solutions.
%28a%2B7%29=%281%2Bsqrt%282a%29%29%5E2
a%2B7=%281%2Bsqrt%282a%29%29%2A%281%2Bsqrt%282a%29%29
a%2B7=1%2B2%2Asqrt%282a%29%2B2a
-a%2B6=2%2Asqrt%282a%29
Square both sides again:
highlight%28%28-a%2B6%29%5E2=%282%2Asqrt%282a%29%29%5E2%29
You should be able to now see that by squaring both sides this second time, we have introduced a potentially incorrect solution to our original problem. We'll have to be alert for this later.
a%5E2-12a%2B36=4%2A2a
a%5E2-20a%2B36=0
This can be factored as follows:
%28a-18%29%2A%28a-2%29=0
---
So, two possible solutions to our original problem are a=18 and a=2. Let's plug them both into the original problem to see if they are actual solutions:
Checking a=18:
sqrt%2818%2B7%29=1%2Bsqrt%282%2A18%29
sqrt%2825%29=1%2Bsqrt%2836%29
5=7.
Obviously a=18 is not a correct solution and was introduced when we squared both sides for a second time. In other words, while a=18 is a solution to our intermediate equation: %28-a%2B6%29%5E2=%282%2Asqrt%282a%29%29%5E2, it is not a solution to our original problem: sqrt%28a%2B7%29=1%2Bsqrt%282a%29.
Checking a=2:
sqrt%282%2B7%29=1%2Bsqrt%282%2A2%29
3=3. This works, so a=2 is the correct solution.