What are the solutions to the equation 2k^8(10-k)=2k^8 ?

The solutions to the equation 2k^8(10-k)=2k^8 are k=0,k=9

Find the zeros of 2k^8(10-k)=2k^8

The zeros of 2k^8(10-k)=2k^8 are k=0,k=9

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2 k^{8} (10 - k) = 2 k^{8}

k = 0, k = 9

Solution steps

2 k^{8} (10 - k) = 2 k^{8}

Move

2 k^{8}

to the left side

2 k^{8} (10 - k) - 2 k^{8} = 0

Factor

2 k^{8} (10 - k) - 2 k^{8} : 2 k^{8} (- k + 9)

2 k^{8} (- k + 9) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

k = 0 or - k + 9 = 0

Solve

- k + 9 = 0 : k = 9

The solutions are

k = 0, k = 9

x^{4} - 29 x + 100 = 0

x, x^{4} - y^{4} = 4

x^{4} = 24

x^{3} + 7 x^{2} + 3 x + 8 = 0

\frac{y ( y - 1 ) ^{2}}{1.5} = 1

What are the solutions to the equation 2k^8(10-k)=2k^8 ?
The solutions to the equation 2k^8(10-k)=2k^8 are k=0,k=9
Find the zeros of 2k^8(10-k)=2k^8
The zeros of 2k^8(10-k)=2k^8 are k=0,k=9