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Question 737371: Please help me solve
(4x)^2 + sqrt(2x+3) = 8x + 1
Thanks a lot, i'm really appreciate your help!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (4x)^2 + sqrt(2x+3) = 8x + 1
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sqrt(2x+3) = -4x^2+8x+1
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2x+3 = (-4x^2+8x+1)(-4x^2+8x+1)
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2x + 3 = 16x^4-32x^3-4x^2 -32x^3+64x^2 + 8x - 4x^2+8x+1
2x+3 = 16x^4 - 64x^3 + 56x^2 + 16x + 1
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16x^4 - 64x^3 + 56x^2 + 14x -2 = 0
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8x^4 - 32x^3 + 28x^2 + 7x - 1 = 0
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I graphed it and found zeroes at
x = -0.28.., x = +0.104.., x = 1.78.. and x = 2.40...
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Check for extraneous zeroes.
Cheers,
Stan H.
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