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Square roots and Quad Form

  • Thread starter imdapolak
  • Start date
1. Sqrt(x)+1=-2Sqrt(x-3)
1.) Sqrt(x)=-2Sqrt(x-3)-1 ( )^2 gives

2.) x= (-2sqrt(x-3)-1)^2 and here I think you need to attempt to foil but im not sure how it works.
(-2sqrt(x-3)-1)(-2sqrt(x-3)-1)=x

3.4sqrt(x-3)+2sqrt(x-3)+2sqrt(x-3)+1=x? Not really sure if this is correct



2. Homework Equations



3. First step I subtract 1 from ea side, and take the whole equation to the power of ^2.
I know for the end of the problem I need to summarize like terms to one side and set it =0 and take the quadratic formula to get 2 answers. 1 will be extraneous.
Any help is greatly appreciated
 

HallsofIvy

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1. Sqrt(x)+1=-2Sqrt(x-3)
1.) Sqrt(x)=-2Sqrt(x-3)-1 ( )^2 gives

2.) x= (-2sqrt(x-3)-1)^2 and here I think you need to attempt to foil but im not sure how it works.
(-2sqrt(x-3)-1)(-2sqrt(x-3)-1)=x

3.4sqrt(x-3)+2sqrt(x-3)+2sqrt(x-3)+1=x? Not really sure if this is correct

Didn't you forget to multiply the squareroots? (That's the "F" in "FOIL".)
[itex]4(\sqrt{x-3})^2+ 4\sqrt{x-3}+ 1= x[/itex]
[itex]4(x-3)+ 4\sqrt{x-3}+ 1= x[/itex]
[itex]4x- 12+ 4\sqrt{x-3}+ 1= x[/itex]
[itex]4\sqrt{x-3}= -3x+ 7[/itex]
Now square again to get rid of that square root.





2. Homework Equations



3. First step I subtract 1 from ea side, and take the whole equation to the power of ^2.
I know for the end of the problem I need to summarize like terms to one side and set it =0 and take the quadratic formula to get 2 answers. 1 will be extraneous.
Any help is greatly appreciated
 
Ok, I think I understood the original problem correctly.
Here is how I read it:
[tex]\sqrt{x}[/tex] + 1= -2[tex]\sqrt{x-3}[/tex]
First thing I would do is square the entire equation. Giving:
([tex]\sqrt{x}[/tex] + 1)[tex]^{2}[/tex] = 4(x-3)
By using the foil method on the left side of the equation and distributing the 4 on the right, you get:
x + 2[tex]\sqrt{x}[/tex] + 1 = 4x - 12
Move the x and the 1 to the right side:
2[tex]\sqrt{x}[/tex] = 3x - 13
Now you can square again and get rid of the last square root:
4x = (3x - 13)[tex]^{2}[/tex]
Foil that:
4x = 9x[tex]^{2}[/tex] - 78x + 169
Move the 4x over:
0 = 9x[tex]^{2}[/tex] - 82x + 169
Then you're ready to use the quadratic formula!

Using the method that you originally attempted would work, but foiling with a square root can get confusing. You always have to square the equation twice to get rid of all of the radicals. For me it is easier to foil a square root when there is not a coefficient in front of it, otherwise I end up forgetting to multiply something along the way. Hope this helps!!!
 

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