Simple Algebra Problem: Finding the Product of Consecutive Even Integers

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kuahji
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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 :-p (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.
 
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kuahji said:
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 :-p (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.

ok so you have
[tex]x(x+2) = 48[/tex]
[tex]x^2 +2x = 48[/tex]

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

then use the quadratic equation to solve for x, and take the solution that fits the problem's description.
 
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d_leet said:
ok so you have
[tex]x(x+2) = 48<br /> x^2 +2x = 48[/tex]
The thing is that you can't just divide by 2x and get [tex]x^2 = 24[/tex]
rewrite the equation as
[tex]x^2 + 2x - 48 = 0[/tex]
then use the quadratic equation to solve for x, and take the solution that fits the problem's description.

Ok, brilliant. Thanks much for your help. It is greatly appreciated :cool:. The algebra rules are slowly coming back to me.
 
Yeah, if you divide [tex]x^2 + 2x = 48[/tex] by 2x then you get:

[tex]\frac {1}{2}x + 1 = \frac {24}{x}[/tex]

Doesn't really get you anywhere. You have to factor to:
[tex](x-6)(x+8) = 0[/tex]
 
One more way. x^2+2x=(x+1)^2-1. So, (x+1)^2=49 So, x+1=(+/-)7