What are the solutions to the equation x^3+4x^2=38x ?

The solutions to the equation x^3+4x^2=38x are x=0,x=-2+sqrt(42),x=-2-sqrt(42)

Find the zeros of x^3+4x^2=38x

The zeros of x^3+4x^2=38x are x=0,x=-2+sqrt(42),x=-2-sqrt(42)

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x^{3} + 4 x^{2} = 38 x

x = 0, x = - 2 + 42, x = - 2 - 42

Decimal Notation

x = 0, x = 4.48074 \dots, x = - 8.48074 \dots

Solution steps

x^{3} + 4 x^{2} = 38 x

Move

38 x

to the left side

x^{3} + 4 x^{2} - 38 x = 0

Factor

x^{3} + 4 x^{2} - 38 x : x (x^{2} + 4 x - 38)

x (x^{2} + 4 x - 38) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x^{2} + 4 x - 38 = 0

Solve

x^{2} + 4 x - 38 = 0 : x = - 2 + 42, x = - 2 - 42

The solutions are

x = 0, x = - 2 + 42, x = - 2 - 42

t, F d^{2} + 3 = t^{3}

21 x^{2} - 18 x - 3 x^{3} = 0

150 x^{2} + 0.38 x^{3} - 80 = 0

x^{3} - 12 x^{2} + 32 x = 0

8 x^{6} - 7 x^{2} = 0

What are the solutions to the equation x^3+4x^2=38x ?
The solutions to the equation x^3+4x^2=38x are x=0,x=-2+sqrt(42),x=-2-sqrt(42)
Find the zeros of x^3+4x^2=38x
The zeros of x^3+4x^2=38x are x=0,x=-2+sqrt(42),x=-2-sqrt(42)