Simple Algebra Problem: Finding the Product of Consecutive Even Integers

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Homework Help Overview

The problem context involves finding the product of two consecutive even integers, specifically stated as "The product of two consecutive even integers is 48." The original poster attempts to set up the equation x(x+2)=48 and is seeking clarification on their approach to solving it.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the equation and the subsequent steps taken by the original poster. There is questioning of the validity of dividing by 2x and suggestions to rewrite the equation in standard form. Some participants propose using the quadratic formula or factoring as alternative methods to solve for x.

Discussion Status

Several participants have provided guidance on how to approach the problem, including rewriting the equation and using the quadratic formula. There is an acknowledgment of different methods being explored, but no explicit consensus has been reached on a single approach.

Contextual Notes

Participants note the original poster's struggle with algebra after a break from school, which may influence their understanding of the problem-solving process. There is also a mention of the need to find solutions that fit the problem's description.

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
 

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