# Homework Help: Simple Algebra Problem

1. Jan 3, 2006

### kuahji

Ok, I have a simple algebra problem I need help with.

The problem states "The product of two consecutive even integers is 48."
So I wrote the equation out as x(x+2)=48. If I factor that I should get x^2+2x=48? Then divide 48 by 2x & get 24. Then squard root of 24? What step am I doing incorrect? I've been racking my brain on this problem for like a half an hour :tongue: (this is what I get for being out of school for 4 years). I mean the answer is simple 6 x 8 = 48. But, I would like to know the steps. Any help is appreciated.

Last edited: Jan 3, 2006
2. Jan 3, 2006

### d_leet

ok so you have
$$x(x+2) = 48$$
$$x^2 +2x = 48$$

The thing is that you can't just divide by 2x and get $$x^2 = 24$$
rewrite the equation as
$$x^2 + 2x - 48 = 0$$

then use the quadratic equation to solve for x, and take the solution that fits the problem's description.

Last edited by a moderator: Jan 3, 2006
3. Jan 3, 2006

### kuahji

Ok, brilliant. Thanks much for your help. It is greatly appreciated . The algebra rules are slowly coming back to me.

4. Jan 4, 2006

### Mindscrape

Yeah, if you divide $$x^2 + 2x = 48$$ by 2x then you get:

$$\frac {1}{2}x + 1 = \frac {24}{x}$$

Doesn't really get you anywhere. You have to factor to:
$$(x-6)(x+8) = 0$$

5. Jan 5, 2006

One more way. x^2+2x=(x+1)^2-1. So, (x+1)^2=49 So, x+1=(+/-)7