# Homework Help: Finding the perfect square

1. Mar 10, 2012

### Miike012

how do you find the perfect square of say

ax2 + b/x2 + c
??

2. Mar 10, 2012

### Mentallic

You can't find the perfect square of that problem precisely, unless you're satisfied with $$\left(\sqrt{ax^2+\frac{b}{x^2}+c}\right)^2$$ which I doubt since it's trivial, but take a look at the expansion of

$$\left(x+\frac{1}{x}\right)^2$$

3. Mar 11, 2012

### Staff: Mentor

You didn't by chance mean (ax2 + b)/(x2 + c), did you? If so, the lack of parentheses around the numerator and denominator completely confused Mentallic about what you're asking.

4. Mar 11, 2012

### Mentallic

That possibility completely skipped my mind

5. Mar 11, 2012

### Staff: Mentor

Mentallic,
Well, I'm about as puzzled by this problem as you must be. The way I read it, the OP just wants to square the original expression, whatever it is.

6. Mar 11, 2012

### Miike012

Nope that is what I ment to say.. I added an example to the paint doc and highlighted the portion in red.

It has to do with finding the surface area of a curve... and basically I was unaware the equation could be turned into a perfect square... so was wondering if there is some pattern I should look for ?

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7. Mar 11, 2012

### Mentallic

I didn't have any doubts about what the OP is trying to do, just what the expression was meant to be once you raised the point.

Mike, like I was saying it doesn't work in general that ax2 + b/x2 + c can be turned into a perfect square, but in this case c happened to be the right number for the job.

When you get to the expression

$$\frac{25}{36}x^8+\frac{1}{2}+\frac{9}{100}x^{-8}$$

You should realize that it could be of the form $$\left(ax^4+bx^{-4}\right)^2$$ where in this case $$a=\sqrt{\frac{25}{36}}=\frac{5}{6}$$
$$b=\sqrt{\frac{9}{100}}=\frac{3}{10}$$

And all you'd need to do is check to see if $$2\cdot \frac{5}{6}\cdot \frac{3}{10} =\frac{1}{2}$$