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Question 46750: sqrt(x+3)+ sqrt(2x-3)=6
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website!
___ ____
Öx+3 + Ö2x-3 = 6
Isolate either radical term
____ ___
Ö2x-3 = 6 - Öx+3
Square both sides:
____ ___
(Ö2x-3)² = (6 - Öx+3)²
It's easy to square the left side, for the squaring
just cancels the square root. However it's not so
easy to square the right side. Put it down twice
and use FOIL
___ ___
2x-3 = (6 - Öx+3)(6 - Öx+3)
On the right:
"F" = 6·6 = 36
___ ___
"O" = (6)(-Öx+3) = -6Öx+3
___ ___
"I" = (-Öx+3)(6) = -6Öx+3
___ ___ ___
"L" = (-Öx+3)(-Öx+3) = (-Öx+3)² = x+3
So we have:
___ ___
2x-3 = 36 - 6Öx+3 - 6Öx+3 + x+3
___
2x-3 = 36 - 12Öx+3 + x+3
___
2x-3 = 39 - 12Öx+3 + x
Isolate the radical term
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12Öx+3 = 39 + x - 2x + 3
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12Öx+3 = 42 - x
Square both sides:
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(12Öx+3)² = (42 - x)²
144(x+3) = (42 - x)(42 - x)
144x + 432 = 1764 - 42x - 42x + x²
144x + 432 = 1764 - 84x + x²
Get 0 on the left:
0 = x² - 228x + 1332
The right side may factor. However since the
numbers are so big, it's probably easier to use
the quadratic formula:
__________________
-(-228) ± Ö(-228)²-4(1)(1332)
x = -------------------------------
2(1)
_____
228 ± Ö46656
x = ----------------
2
228 ± 216
x = -------------
2
Using the +
228 + 216
x = -------------
2
444
x = ----- = 222
2
Using the -
228 - 216
x = -------------
2
12
x = ---- = 6
2
Now we must check, because often we get extraneous
answers in equations with even roots. Checking the
answer x = 222
___ ____
Öx+3 + Ö2x-3 = 6
_____ ________
Ö222+3 + Ö2(222)-3 = 6
___ ___
Ö225 + Ö441 = 6
15 + 21 = 6
36 = 6
No that doesn't check. So we discard that answer.
Checking the answer x=6
___ ____
Öx+3 + Ö2x-3 = 6
___ ______
Ö6+3 + Ö2(6)-3 = 6
_ _
Ö9 + Ö9 = 6
3 + 3 = 6
6 = 6
That checks. So there is one solution,
x = 6
Edwin
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