SOLUTION: this is from the 10th edition of the beginning algebra book. There are no instructions other than to solve. {{{ sqrt(3x+3)+sqrt(x+2)=5 }}}

Algebra ->  Radicals -> SOLUTION: this is from the 10th edition of the beginning algebra book. There are no instructions other than to solve. {{{ sqrt(3x+3)+sqrt(x+2)=5 }}}      Log On


   

Question 460650: this is from the 10th edition of the beginning algebra book. There are no instructions other than to solve.
+sqrt%283x%2B3%29%2Bsqrt%28x%2B2%29=5+
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
+sqrt%283x%2B3%29%2Bsqrt%28x%2B2%29=5+

Let u = one of the radical terms, say

u=sqrt%28x%2B2%29

+sqrt%283x%2B3%29%2Bu=5+

Isolate the radical term:

+sqrt%283x%2B3%29=5-u

Square both sides:

3x%2B3+=+25-10u%2Bu%5E2

Replace u by sqrt%28x%2B2%29 and u² by x+2

3x%2B3+=+25-10sqrt%28x%2B2%29%2Bx%2B2

3x%2B3+=+27-10sqrt%28x%2B2%29%2Bx

Isolate the radical term:

10sqrt%28x%2B2%29=24-2x

Divide every term by 2 since they are all even:

5sqrt%28x%2B2%29=12-x

Square both sides:

25%28x%2B2%29+=+144-24x%2Bx%5E2

25x + 50 = 144 - 24x + x²

Get 0 on the left side:

0 = x² - 49x + 94

Factor the right side:

0 = (x - 2)(x - 47}

Use the zero factor principle

x - 2 = 0        x - 47 = 0
    x = 2             x = 47

But we must always check a radical equation,
because sometimes there may be a "phony"
solution, called "extraneous".  

Checking x = 2

+sqrt%283x%2B3%29%2Bsqrt%28x%2B2%29=5+
+sqrt%283%282%29%2B3%29%2Bsqrt%282%2B2%29=5+
+sqrt%286%2B3%29%2Bsqrt%282%2B2%29=5+
+sqrt%289%29%2Bsqrt%284%29=5+
3%2B2=5
5=5

That checks so x = 2 is a solution.

Checking x = 47

+sqrt%283x%2B3%29%2Bsqrt%28x%2B2%29=5+
+sqrt%283%2847%29%2B3%29%2Bsqrt%2847%2B2%29=5+
+sqrt%28141%2B3%29%2Bsqrt%2849%29=5+
+sqrt%28144%29%2B7=5+
12%2B7=5
19=5

That is false so x = 47 is a "phony"
or "irrational" solution.  So the
only solution is x = 2

Edwin