What are the solutions to the equation 20x=x^2(x)+36x-2x(x)(x) ?

The solutions to the equation 20x=x^2(x)+36x-2x(x)(x) are x=0,x=-4,x=4

Find the zeros of 20x=x^2(x)+36x-2x(x)(x)

The zeros of 20x=x^2(x)+36x-2x(x)(x) are x=0,x=-4,x=4

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20 x = x^{2} (x) + 36 x - 2 x (x) (x)

x = 0, x = - 4, x = 4

Solution steps

20 x = x^{2} (x) + 36 x - 2 x (x) (x)

Expand

x^{2} (x) + 36 x - 2 x (x) (x) : - x^{3} + 36 x

20 x = - x^{3} + 36 x

Switch sides

- x^{3} + 36 x = 20 x

Move

20 x

to the left side

- x^{3} + 16 x = 0

Factor

- x^{3} + 16 x : - x (x + 4) (x - 4)

- x (x + 4) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 4 = 0 or x - 4 = 0

Solve

x + 4 = 0 : x = - 4

Solve

x - 4 = 0 : x = 4

The solutions are

x = 0, x = - 4, x = 4

y, y^{2} - y^{4} = x

x^{4} - 3 x + 4 = 0

2 x^{3} - 3 x^{2} - 2 x + 3 = 0

\frac{1}{3} x^{4} = 27

(1 + x)^{25} = 4.29

What are the solutions to the equation 20x=x^2(x)+36x-2x(x)(x) ?
The solutions to the equation 20x=x^2(x)+36x-2x(x)(x) are x=0,x=-4,x=4
Find the zeros of 20x=x^2(x)+36x-2x(x)(x)
The zeros of 20x=x^2(x)+36x-2x(x)(x) are x=0,x=-4,x=4