Simple. (Problem 8 8 )

Algebra Level pending

( 2 x + 1 ) ! = x ( 0 ) = ( x ) 0 (2x + 1)! = x^{(0)} = (x)_0

Find the negative solution of x x .

Note:

x ( 0 ) x^{(0)} denotes the rising factorial

( x ) 0 (x)_0 denotes the falling factorial

Tip: If you want, you can remove the ! ! sign.

For more information about rising and falling factorials, check out the Wikipedia article .


The answer is -0.5.

Yajat Shamji by Yajat Shamji , 16, United Kingdom

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1 solution

Yajat Shamji
Jul 12, 2020

Note that:

x ( 0 ) = ( x ) 0 = 1 x^{(0)} = (x)_0 = 1

Therefore:

( 2 x + 1 ) ! = 1 (2x + 1)! = 1

Now:

0 ! = 1 0! = 1

1 ! = 1 1! = 1

Therefore, after removing the factorial sign and solving, x = 0 x = 0 or x = − x = - 1 2 \frac{1}{2}

Since we're asked to find the negative solution of x x , x = − x = - 1 2 \frac{1}{2}

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