SOLUTION: Can I get your help? My problem is this: {{{3x=1+sqrt(4x^2+x+7)}}} I started by subtracting the 1 from both sides which gave me {{{3x-1=sqrt(4x^2+x+7)}}} Now, would I add 4x^

Algebra ->  Radicals -> SOLUTION: Can I get your help? My problem is this: {{{3x=1+sqrt(4x^2+x+7)}}} I started by subtracting the 1 from both sides which gave me {{{3x-1=sqrt(4x^2+x+7)}}} Now, would I add 4x^      Log On


   

Question 992872: Can I get your help? My problem is this:
3x=1%2Bsqrt%284x%5E2%2Bx%2B7%29
I started by subtracting the 1 from both sides which gave me 3x-1=sqrt%284x%5E2%2Bx%2B7%29 Now, would I add 4x^2 and x which would give me sqrt+%285x%5E3%2B7%29Now my equation looks like this: +3X-1=SQRT%285x%5E3%2B7%29 From here I would square both sides %283x-1%29%5E2=%28sqrt+5x%5E3%2B7%29%5E2
I get 9x^2+3x-1-3x+1-1=5x^3+7. If I continue, 3x-1-3x+1 would cancel out which makes the equation look like this:
9x^2-1=5x^3+7. I subtract 9x^2 from both sides and I subtract 7 from both sides so I end up with -8=-4x. Then I divide by -4 and my answer would be 2. Can you verify to see if I made a mistake?
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Can I get your help? My problem is this:
3x=1%2Bsqrt%284x%5E2%2Bx%2B7%29
I started by subtracting the 1 from both sides which gave me 3x-1=sqrt%284x%5E2%2Bx%2B7%29
Now, would I add 4x^2 and x which would give me sqrt+%285x%5E3%2B7%29 *********
No, you can't do that. You can't add x^2 terms and x terms.
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3x-1=sqrt%284x%5E2%2Bx%2B7%29
Square both sides
9x%5E2+-+6x+%2B+1+=+4x%5E2+%2B+x+%2B+7
etc

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
3x=1+sqrt(4x^2+x+7)
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3x-1 = sqrt(4x^2+x+7)
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Square both sides to get:
9x^2 -6x + 1 = 4x^2 + x + 7
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5x^2 - 7x - 6 = 0
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5x^2 - 10x+3x - 6 = 0
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5x(x-2)+3(x-2) = 0
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(x-2)(5x+3) = 0
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x = 2 or x = -3/5
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Cheers,
Stan H.
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