What Is the Correct Method to Expand a Radical Expression?

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Homework Statement


How do you expand a radical?


Homework Equations





The Attempt at a Solution

 
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You have to rephrase your question. As far as I know, you can only expand a term, e.g. [tex](x - 2)^2[/tex]
 
Sorry I meant a term. like you example you just gave. I am confused on how to expand them for doing radical equations.
 
[tex](x - \sqrt{2})^2 = (x - \sqrt{2})(x - \sqrt{2})[/tex]

Do you know how to F.O.I.L.? (multiply first by outer and inner by last)

[tex]x^2- x\sqrt{2} - x\sqrt{2} + 2[/tex]

Or:

[tex](a + b)^2 = a^2 + 2ab + b^2[/tex]

Just take it slow. You can always check your work with a calculator.
 
Oh so you multiply from inner to outer. I get it now.
 
[tex]\sqrt{2x*2}[/tex]+[tex]\sqrt{3x}[/tex]=22

I am having trouble with this problem, I used the quadratic formual and the binomial square formula. I got x=9.75561, I just checked it and its wrong. I think the problem is when i try to exand [tex]\sqrt{2x*2}[/tex], after the expansion of the right side i got 484-4[tex]^{x}_{2}[/tex]-8x+4.
 
Stratosphere said:
[tex]\sqrt{2x*2}[/tex]+[tex]\sqrt{3x}[/tex]=22
I'm having trouble understanding your notation. Is that a 2x TIMES 2 inside the first square root? Can you clean that up?


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Sorry, I made a typo it was suppose to be +.
 
You can't expand [tex]\sqrt{2x + 2}[/tex], because it isn't a term raised to a power. You understand me?

Say you had [tex](\sqrt{2x + 2})^2[/tex], you could expand that, because it's a term raised to a power.

If your question is [tex]\sqrt{2x + 2} + \sqrt{3x} = 22[/tex] , then try squaring both sides, and see what happens.
 
If I only square each side and leave the to square roots on the left alone the when I comebine the 3x and the 2x i get 5x+2=484, then i sybract the 2 to get 5x=482, I then do the divition and I get 96.4. That's not right.
 
Oh, yeah, move the [tex]\sqrt{3x}[/tex] to the right hand side. Then square each side.
 
You're going to have the term [tex](22 - \sqrt{3x})[/tex] to expand on the RHS.
 
General_Sax said:
You can't expand [tex]\sqrt{2x + 2}[/tex], because it isn't a term raised to a power. You understand me?

No sorry, I don't understand you..

[tex]\sqrt{a}=a^{\frac{1}{2}}[/tex] and that's a term raised to a power. It's just not an integral power.

[tex]\sqrt{2x+2}=(2x+2)^{\frac{1}{2}}[/tex]
 
Mentallic said:
No sorry, I don't understand you..

[tex]\sqrt{a}=a^{\frac{1}{2}}[/tex] and that's a term raised to a power. It's just not an integral power.

[tex]\sqrt{2x+2}=(2x+2)^{\frac{1}{2}}[/tex]

Can you expand [tex]\sqrt{2x+2}=(2x+2)^{\frac{1}{2}}[/tex] ? I'm sorry I wasn't more accurate, but I was just trying to help this person with their homework. If by some chance I came off as snobby(or whatever), I didn't intend for it.
 
Stratosphere said:
If I only square each side and leave the to square roots on the left alone the when I comebine the 3x and the 2x i get 5x+2=484, then i sybract the 2 to get 5x=482, I then do the divition and I get 96.4. That's not right.
You're right, that's incorrect. So you tried to do
[tex](\sqrt{2x + 2} + \sqrt{3x})^{2}[/tex]
and you got
[tex]2x + 2 + 3x[/tex]
which is incorrect. Squaring that radical expression doesn't work that way.

Try what General Sax suggested, that is, first subtract the [tex]\sqrt{3x}[/tex] from both sides to get
[tex]\sqrt{2x + 2} = 22 - \sqrt{3x}[/tex]

Now square both sides.
[tex](\sqrt{2x + 2})^{2} = (22 - \sqrt{3x})^{2}[/tex]
In this case the left side becomes 2x + 2. You'll need to use the special product property [tex](a - b)^{2}=a^{2} - 2ab + b^{2}[/tex] to expand the right side. Can you take it from there?


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