HI !!
How can I solve this equation ???
$\displaystyle (x+1)(x+2)(x+3)(x+4) = 120 \ \ \ \ \ \ $
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
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} $
Hello, yehoram!
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: .$\displaystyle \{2,3,4,5\}\,\text{ and }\,\{\text{-}5,\text{-}4,\text{-}3,\text{-}2\}$