how do you solve this :(

i wasnt at school and i have exam tommorow :(

x!/(x-2)! + (x-1)!/(x-3)! = 8

Guest Jun 5, 2017

#3**+1 **

x! Is x*(x-1)*(x-2)*...*2*1

That means that x!/(x-2)! → x*(x-1)*(x-2)*..,*2*1/( (x-2)*(x-3)*...*2*1 ) → x*(x-1) as everything else cancels out.

Similarly (x-1)!/(x-3)! → (x-1)*(x-2)*(x-3)*...*2*1/( (x-3)*...*2*1 ) → (x-1)*(x-2)

Hence: x!/(x-2)! + (x-1)!/(x-3)! = 8 becomes: x*(x-1) + (x-1)*(x-2) = 8

Expanding and simplifying this we get: 2x^{2} - 4x + 2 = 8 or x^{2} - 2x -3 = 0

This factors as: (x + 1)(x - 3) = 0

Becase the question originally involved factorials we can only accept a positive result, so x = 3

Check:

3!/(3-2)! = 6/1 → 6

(3-1)!/(3-3)! = 2/1 → 2. (Note that 0! = 1)

These two sum to 8.

Alan
Jun 5, 2017

#1**0 **

x!/(x-2)! + (x-1)!/(x-3)! = 8 is the same as (x - 2) (x - 1) + x (x - 1) = 8

Solve for x:

(x - 2) (x - 1) + x (x - 1) = 8

Expand out terms of the left hand side:

2 x^2 - 4 x + 2 = 8

Divide both sides by 2:

x^2 - 2 x + 1 = 4

Write the left hand side as a square:

(x - 1)^2 = 4

Take the square root of both sides:

x - 1 = 2 or x - 1 = -2

Add 1 to both sides:

x = 3 or x - 1 = -2

Add 1 to both sides:

**Answer: | x = 3 or x = -1**

Guest Jun 5, 2017

#3**+1 **

Best Answer

x! Is x*(x-1)*(x-2)*...*2*1

That means that x!/(x-2)! → x*(x-1)*(x-2)*..,*2*1/( (x-2)*(x-3)*...*2*1 ) → x*(x-1) as everything else cancels out.

Similarly (x-1)!/(x-3)! → (x-1)*(x-2)*(x-3)*...*2*1/( (x-3)*...*2*1 ) → (x-1)*(x-2)

Hence: x!/(x-2)! + (x-1)!/(x-3)! = 8 becomes: x*(x-1) + (x-1)*(x-2) = 8

Expanding and simplifying this we get: 2x^{2} - 4x + 2 = 8 or x^{2} - 2x -3 = 0

This factors as: (x + 1)(x - 3) = 0

Becase the question originally involved factorials we can only accept a positive result, so x = 3

Check:

3!/(3-2)! = 6/1 → 6

(3-1)!/(3-3)! = 2/1 → 2. (Note that 0! = 1)

These two sum to 8.

Alan
Jun 5, 2017