What are the solutions to the equation g(n)=n^3-2n,g(n+4) ?

The solutions to the equation g(n)=n^3-2n,g(n+4) are n=0,n=sqrt(g+2),n=-sqrt(g+2)

Find the zeros of g(n)=n^3-2n,g(n+4)

The zeros of g(n)=n^3-2n,g(n+4) are n=0,n=sqrt(g+2),n=-sqrt(g+2)

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g (n) = n^{3} - 2 n, g (n + 4)

n = 0, n = g + 2, n = - g + 2

Solution steps

g (n) = n^{3} - 2 n, g (n + 4)

Refine

g n = n^{3} - 2 n

Switch sides

n^{3} - 2 n = g n

Move

g n

to the left side

n^{3} - 2 n - g n = 0

Factor

n^{3} - 2 n - g n : n (n^{2} - 2 - g)

n (n^{2} - 2 - g) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n = 0 or n^{2} - 2 - g = 0

Solve

n^{2} - 2 - g = 0 : n = g + 2, n = - g + 2

The solutions are

n = 0, n = g + 2, n = - g + 2

a^{4} - 8 a^{2} = - 15

x^{4} - 8 x^{2} = 0

t^{7} = 5^{5}

6 x^{3} - 8 = 46

6 x^{3} - 8 = 40

What are the solutions to the equation g(n)=n^3-2n,g(n+4) ?
The solutions to the equation g(n)=n^3-2n,g(n+4) are n=0,n=sqrt(g+2),n=-sqrt(g+2)
Find the zeros of g(n)=n^3-2n,g(n+4)
The zeros of g(n)=n^3-2n,g(n+4) are n=0,n=sqrt(g+2),n=-sqrt(g+2)