How do I solve a quadratic equation involving square roots without foiling?

AI Thread Summary
To solve the quadratic equation involving square roots, the initial step is to isolate the square root and square both sides of the equation. This leads to a new equation where further simplification and squaring may be necessary to eliminate remaining square roots. The discussion emphasizes the importance of using the FOIL method correctly when squaring binomials. After simplifying, the equation can be rearranged into standard quadratic form, allowing the quadratic formula to be applied for finding solutions. Ultimately, this process may yield extraneous solutions that need to be checked against the original equation.
imdapolak
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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 I am 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



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
 
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imdapolak said:
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 I am 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".)
4(\sqrt{x-3})^2+ 4\sqrt{x-3}+ 1= x
4(x-3)+ 4\sqrt{x-3}+ 1= x
4x- 12+ 4\sqrt{x-3}+ 1= x
4\sqrt{x-3}= -3x+ 7
Now square again to get rid of that square root.





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:
\sqrt{x} + 1= -2\sqrt{x-3}
First thing I would do is square the entire equation. Giving:
(\sqrt{x} + 1)^{2} = 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\sqrt{x} + 1 = 4x - 12
Move the x and the 1 to the right side:
2\sqrt{x} = 3x - 13
Now you can square again and get rid of the last square root:
4x = (3x - 13)^{2}
Foil that:
4x = 9x^{2} - 78x + 169
Move the 4x over:
0 = 9x^{2} - 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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