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

In summary, the conversation involves solving an equation with square roots on both sides. The first step is to square the entire equation, then use the foil method to expand the left side and distribute the coefficient on the right side. This leads to a quadratic equation, which can be solved using the quadratic formula. Alternatively, one can also foil the square root and then square the equation again to get rid of all radicals.
  • #1
imdapolak
10
0
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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  • #2
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".)
[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.





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
 
  • #3
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!
 

What is a square root?

A square root is a number that, when multiplied by itself, yields the original number. For example, the square root of 25 is 5 because 5 multiplied by itself equals 25.

How do you find the square root of a number?

The most common method for finding a square root is by using a calculator. However, you can also use long division or a method called "prime factorization" to find the square root of a number.

What is the quadratic formula?

The quadratic formula is a mathematical formula used to solve quadratic equations, which are equations in the form of ax^2 + bx + c = 0. The formula is: x = (-b ± √(b^2 - 4ac)) / 2a

How is the quadratic formula derived?

The quadratic formula is derived from completing the square on a general quadratic equation. This process involves manipulating the equation to create a perfect square trinomial, which can then be easily solved using the square root method.

What is the relationship between square roots and quadratic equations?

Square roots and quadratic equations are closely related because the solutions to a quadratic equation are often found by taking the square root of both sides of the equation. In some cases, the solutions may be irrational numbers (numbers that cannot be expressed as a fraction) which require the use of the quadratic formula to find.

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