Good job.Originally Posted byChester

Find some factors of 168: 12 and 14

which diffrence is 2 thus,

Thus,

Results 1 to 7 of 7

- July 30th 2006, 02:38 PM #1

- Joined
- Jul 2006
- Posts
- 25

## Help again

Please explain how to solve this.

The product of two consecutive even integers is 168. Find the two integers.

Hint: Assume the first integer is x

Set up the equation first that satisfies the given condition. Then solve the equation.

This is the equation that i came up with. Ican t seem to get past

x(x+2)=168

x^2+2x-168=0 (i cant get past here)

Thanks,

Chester

- July 30th 2006, 02:40 PM #2

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- July 30th 2006, 02:44 PM #3

- July 30th 2006, 02:59 PM #4

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,024
- Thanks
- 832

Hello, Chester!

You did an excellent on job on the hard part: setting up the equation.

So where is your difficulty?

I assume you know to solve a quadratic equation

. . and that you know how to factor.

Did you run of factorings to try?

We want two numbers with a product of 168 and a difference of 2.

Start dividing 168 by 1, 2, 3, . . .

There! a difference of 2

Therefore: .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Of course, you can use the*Quadratic Formula*for factoring . . .

- July 30th 2006, 03:18 PM #5

- Joined
- Jul 2006
- Posts
- 25

- July 30th 2006, 06:10 PM #6

- July 30th 2006, 06:18 PM #7

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10