# How to expand a radical

1. Mar 25, 2009

### Stratosphere

1. The problem statement, all variables and given/known data
How do you expand a radical?

2. Relevant equations

3. The attempt at a solution

2. Mar 25, 2009

### General_Sax

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

3. Mar 25, 2009

### Stratosphere

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

4. Mar 25, 2009

### 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.

5. Mar 25, 2009

### Stratosphere

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

6. Mar 26, 2009

### 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 expantion of the right side i got 484-4$$^{x}_{2}$$-8x+4.

7. Mar 26, 2009

### HallsofIvy

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

8. Mar 26, 2009

### yeongil

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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9. Mar 26, 2009

### Stratosphere

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

10. Mar 26, 2009

### 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.

11. Mar 26, 2009

### 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.

12. Mar 26, 2009

### General_Sax

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

13. Mar 26, 2009

### General_Sax

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

14. Mar 27, 2009

### Mentallic

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.

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

15. Mar 27, 2009

### General_Sax

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.

16. Mar 27, 2009

### yeongil

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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