how do you solve this :(
i wasnt at school and i have exam tommorow :(
x!/(x-2)! + (x-1)!/(x-3)! = 8
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: 2x2 - 4x + 2 = 8 or x2 - 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.
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
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: 2x2 - 4x + 2 = 8 or x2 - 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.