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# how do you solve this :(

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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
+27057
+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.

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

x = 3 or x - 1 = -2

Answer: | x = 3        or         x = -1

Guest Jun 5, 2017
#3
+27057
+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.

Alan  Jun 5, 2017
#4
0

<3 awesome

Guest Jun 5, 2017