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Question 76611: can you help me solve this problem
square root of 2x + 8 - square root of 2x + 1 = 1
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! :
square root of 2x + 8 - square root of 2x + 1 = 1
Here it is step-by-step
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Sqrt(2x+8) - Sqrt(2x+1) = 1
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Sqrt(2x+8) = Sqrt(2x+1) + 1; added Sqrt(2x+1) to both sides
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Square both sides, gets rid of the radical on the left:
FOIL (Sqrt(2x+1)+ 1)*(Sqrt(2x+1) + 1) on the right
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2x + 8 = (2x+1) + 2Sqrt(2x+1) + 1
:
2x + 8 = 2x + 1 + 1 + 2Sqrt(2x+1)
2x + 8 = 2x + 2 + 2Sqrt(2x+1)
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Subtract 2x and 2 from both sides, isolates the radical on the right
2x - 2x + 8 - 2 = 2Sqrt(2x+1)
6 = 2Sqrt(2x + 1)
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Square both sides, gets rid of the 2nd radical, you have:
36 = 4(2x + 1)
36 = 8x + 4; multiplied what's inside the brackets
36 - 4 = 8x: subtract 4 from both sides
32 = 8x
x = 32/8
x = 4
:
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Check solution in original equation:
Sqrt(2(4) + 8) - Sqrt(2(4) + 1) = 1
Sqrt(8 + 8) - Sqrt(8 + 1) = 1
Sqrt(16) - Sqrt(9) = 1; proves our solution is correct
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How about this, do you have an idea of how to do these now?
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