HI !!

How can I solve this equation ???

$\displaystyle (x+1)(x+2)(x+3)(x+4) = 120 \ \ \ \ \ \ $

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- Dec 28th 2010, 04:37 AMyehoramhard equation !
HI !!

How can I solve this equation ???

$\displaystyle (x+1)(x+2)(x+3)(x+4) = 120 \ \ \ \ \ \ $ - Dec 28th 2010, 04:50 AMProve It
1. Expand

2. Set equal to 0

3. Factorise if possible and solve using NFL. - Dec 28th 2010, 04:54 AMDrSteve
Here's a clever little shortcut. 120 is equal to 5!. So x = 1 works.

- Dec 28th 2010, 04:55 AMSudharaka
Dear yehoram,

$\displaystyle (x+1)(x+2)(x+3)(x+4) = 120=2\times 3\times 4\times 5$

Therefore it is clear that x=1 is a solution.

$\displaystyle (x+1)(x+2)(x+3)(x+4)-120=(x-1)(Ax^3+Bx^2+Cx+D)=0$

Evaluating A,B,C and D will give you a cubic equation. The solutions can be obtained using several methods such as the Cardano's method. Refer: Cubic function - Wikipedia, the free encyclopedia - Dec 28th 2010, 06:15 AMsimplependulum
The key to this problem is to partition the products into two : (1) the first and the last , (2) the 2nd and the 3rd , ie

$\displaystyle (x+1)(x+2)(x+3)(x+4) = 120 $

$\displaystyle [(x+1)(x+4)][(x+2)(x+3)] = 120 $

$\displaystyle (x^2 + 5x + 4 )(x^2 + 5x + 6) = 120 $

Then sub. $\displaystyle u= x^2 + 5x + 5 $ so we have

$\displaystyle (u-1)(u+1) = 120 $ or

$\displaystyle u^2 = 121 $ thus

$\displaystyle u = 11 $ or $\displaystyle u = -11 $

$\displaystyle x^2 + 5x - 6 = 0 $ or $\displaystyle x^2 + 5x + 16 = 0 $

The four solutions are

$\displaystyle -6,1 , \frac{-5+\sqrt{39} i }{2} , \frac{-5-\sqrt{39} i }{2} $ - Dec 28th 2010, 06:21 AMSoroban
Hello, yehoram!

Quote:

How can I solve this equation?

. . $\displaystyle (x+1)(x+2)(x+3)(x+4) \:=\: 120 $

We have: the product of four "consecutive" numbers equals 120.

The only candidates are: .$\displaystyle \{2,3,4,5\}\,\text{ and }\,\{\text{-}5,\text{-}4,\text{-}3,\text{-}2\}$

- Dec 28th 2010, 07:19 AMArchie Meade
- Dec 28th 2010, 08:12 AMWilmer