What Is the Correct Method to Expand a Radical Expression?

[Stratosphere]

Homework Statement

How do you expand a radical?

Homework Equations

The Attempt at a Solution

 

[General_Sax]

You have to rephrase your question. As far as I know, you can only expand a term, e.g. (x - 2)^2

 

[Stratosphere]

Sorry I meant a term. like you example you just gave. I am confused on how to expand them for doing radical equations.

 

[General_Sax]

(x - \sqrt{2})^2 = (x - \sqrt{2})(x - \sqrt{2})

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

x^2- x\sqrt{2} - x\sqrt{2} + 2

Or:

(a + b)^2 = a^2 + 2ab + b^2

Just take it slow. You can always check your work with a calculator.

 

[Stratosphere]

Oh so you multiply from inner to outer. I get it now.

 

[Stratosphere]

\sqrt{2x*2}+\sqrt{3x}=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 \sqrt{2x*2}, after the expansion of the right side i got 484-4^{x}_{2}-8x+4.

 

[HallsofIvy]

\sqrt{2x*2}? That's just 2\sqrt{x}. Do you mean \sqrt{2x+2}?

 

[yeongil]

\sqrt{2x*2}+\sqrt{3x}=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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[Stratosphere]

Sorry, I made a typo it was suppose to be +.

 

[General_Sax]

You can't expand \sqrt{2x + 2}, because it isn't a term raised to a power. You understand me?

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

If your question is \sqrt{2x + 2} + \sqrt{3x} = 22 , then try squaring both sides, and see what happens.

 

[Stratosphere]

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.

 

[General_Sax]

Oh, yeah, move the \sqrt{3x} to the right hand side. Then square each side.

 

[General_Sax]

You're going to have the term (22 - \sqrt{3x}) to expand on the RHS.

 

[Mentallic]

You can't expand \sqrt{2x + 2}, because it isn't a term raised to a power. You understand me?

No sorry, I don't understand you..

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

\sqrt{2x+2}=(2x+2)^{\frac{1}{2}}

 

[General_Sax]

No sorry, I don't understand you..

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

\sqrt{2x+2}=(2x+2)^{\frac{1}{2}}

Can you expand \sqrt{2x+2}=(2x+2)^{\frac{1}{2}} ? 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.

 

[yeongil]

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
(\sqrt{2x + 2} + \sqrt{3x})^{2}
and you got
2x + 2 + 3x
which is incorrect. Squaring that radical expression doesn't work that way.

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

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


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