What are the solutions to the equation solvefor k,(n-k)(-k)=k(n-k) ?

The solutions to the equation solvefor k,(n-k)(-k)=k(n-k) are k=0,k=n

Find the zeros of solvefor k,(n-k)(-k)=k(n-k)

The zeros of solvefor k,(n-k)(-k)=k(n-k) are k=0,k=n

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solve for

k, (n - k) (- k) = k (n - k)

k = 0, k = n

Solution steps

(n - k) (- k) = k (n - k)

Move

k (n - k)

to the left side

(n - k) (- k) - k (n - k) = 0

Factor

(n - k) (- k) - k (n - k) : - 2 k (n - k)

- 2 k (n - k) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

k = 0 or n - k = 0

Solve

n - k = 0 : k = n

The solutions to the quadratic equation are:

k = 0, k = n

4 x^{2} - 28 x + 57 = 0

- c^{2} + c + 1 = 0

k^{2} - k = 0

(2 y + 6)^{2} = 36

16 x^{2} + 56 x + 49 = 0

What are the solutions to the equation solvefor k,(n-k)(-k)=k(n-k) ?
The solutions to the equation solvefor k,(n-k)(-k)=k(n-k) are k=0,k=n
Find the zeros of solvefor k,(n-k)(-k)=k(n-k)
The zeros of solvefor k,(n-k)(-k)=k(n-k) are k=0,k=n