SOLUTION: Solve 2√(3x + 1) + 3√(4x + 5) = 23

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Question 1156170: Solve
2√(3x + 1) + 3√(4x + 5) = 23

Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

2sqrt%283x+%2B+1%29+%2B+3sqrt%284x+%2B+5%29+=+23........square both sides

%282sqrt%283x+%2B+1%29+%2B+3sqrt%284x+%2B+5%29%29%5E2+=+23%5E2



4%283x+%2B+1%29+%2B12sqrt%283x+%2B+1%29%2Asqrt%284x+%2B+5%29+%2B+9%284x+%2B+5%29+=+529

12x+%2B+4+%2B12sqrt%283x+%2B+1%29%2Asqrt%284x+%2B+5%29+%2B+36x+%2B+45+=+529

12sqrt%283x+%2B+1%29%2Asqrt%284x+%2B+5%29+%2B+48x+%2B+49+=+529

12sqrt%283x+%2B+1%29%2Asqrt%284x+%2B+5%29+=+529-48x-49

12sqrt%283x+%2B+1%29%2Asqrt%284x+%2B+5%29+=+480-48x ......simplify, both sides divide by 12

sqrt%283x+%2B+1%29%2Asqrt%284x+%2B+5%29+=+40-4x .......square both sides

%28sqrt%283x+%2B+1%29%2Asqrt%284x+%2B+5%29%29%5E2+=+%2840-4x%29%5E2

%283x+%2B+1%29%2A%284x+%2B+5%29+=+1600-320x%2B16x%5E2

12x%5E2+%2B+19x+%2B+5+=+1600-320x%2B16x%5E2

0=+1600-320x%2B16x%5E2-12x%5E2+-+19x+-5+

4x%5E2-339x%2B1595=0

%28x+-+5%29+%284x+-+319%29+=+0

solutions:
if x-5=0=>x=5
if %284x+-+319%29+=+0=>4x+=+319=>x+=+319%2F4=>x+=+79.75

only solution that satisfy given equation is x=5



Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
2sqrt%283x+%2B+1%29+%2B+3sqrt%284x+%2B+5%29+=+23

Isolate one of the square root terms, it doesn't matter which one.
I'll isolate the first one:

2sqrt%283x+%2B+1%29+=+23+-+3sqrt%284x+%2B+5%29

Square both sides:

%282sqrt%283x+%2B+1%29%29%5E2+=+%2823+-+3sqrt%284x+%2B+5%29%29%5E2

Square each factor on the left.  Copy the binomial on the right 
twice and multiply using FOIL:



4%283x%2B1%29=23%5E2-69sqrt%284x%2B5%29-69sqrt%284x%2B5%29%2B%283sqrt%284x%2B5%29%29%5E2

12x%2B4=529-138sqrt%284x%2B5%29%2B3%5E2%284x%2B5%29

12x%2B4=529-138sqrt%284x%2B5%29%2B9%284x%2B5%29

12x%2B4=529-138sqrt%284x%2B5%29%2B36x%2B45%29

12x%2B4=574-138sqrt%284x%2B5%29%2B36x%29

12x=570-138sqrt%284x%2B5%29%2B36x%29

It will make things easier to divide each term by 2

6x=285-69sqrt%284x%2B5%29%2B18x

Isolate the square root term on the right:

-12x-285=-69sqrt%284x%2B5%29

Divide through by -1

12x%2B285=69sqrt%284x%2B5%29

It will make things easier to divide each term by 3

4x%2B95=23sqrt%284x%2B5%29

Square both sides:

%284x%2B95%29%5E2=%2823sqrt%284x%2B5%29%29%5E2

%284x%2B95%29%284x%2B95%29=23%5E2%284x%2B5%29

16x%5E2%2B380x%2B380x%2B9025=529%284x%2B5%29

16x%5E2%2B760x%2B9025=2116x%2B2645%29

16x%5E2-1356x%2B6380=0

Divide through by 4

4x%5E2-339x%2B1595=0

%28x-5%29%284x-319%29=0

x-5=0   4x=319
  x=5    x=319%2F4
               x=79.75   

Checking x=5 in the original equation:

2sqrt%283x+%2B+1%29+%2B+3sqrt%284x+%2B+5%29+=+23
2sqrt%283%285%29+%2B+1%29+%2B+3sqrt%284%285%29+%2B+5%29+=+23
2sqrt%2816%29+%2B+3sqrt%2820+%2B+5%29+=+23
2%284%29+%2B+3sqrt%2825%29+=+23
2%284%29+%2B+3%285%29+=+23
8+%2B+15+=+23
23=23

That checks.

Checking x=79.75 in the original equation

2sqrt%283x+%2B+1%29+%2B+3sqrt%284x+%2B+5%29+=+23
2sqrt%283%2879.75%29+%2B+1%29+%2B+3sqrt%284%2879.75%29+%2B+5%29+=+23
2sqrt%28240.25%29+%2B+3sqrt%28324%29+=+23
2%2815.5%29+%2B+3%2818%29+=+23
31%2B54+=+23

[That gives us 85, and so 79.75 is an extraneous solution.
It is interesting that extraneous solutions usually would 
have checked if one of the square root terms could have 
its sign changed in front.  We see here that if the 31 had 
been -31, we'd have had -31+54=23, which would have checked.
This is due to the fact that when we square a negative number,
we get the same thing as when we square the corresponding
positive number.  We squared both sides twice, so we should
expect that.] 

There is only one solution x=5.

Edwin