What are the solutions to the equation 12(x+4)=x^3 ?

The solutions to the equation 12(x+4)=x^3 are x\approx 4.71060…

Find the zeros of 12(x+4)=x^3

The zeros of 12(x+4)=x^3 are x\approx 4.71060…

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12 (x + 4) = x^{3}

x \approx 4.71060 \dots

Solution steps

12 (x + 4) = x^{3}

Expand

12 (x + 4) : 12 x + 48

12 x + 48 = x^{3}

Switch sides

x^{3} = 12 x + 48

Move

48

to the left side

x^{3} - 48 = 12 x

Move

12 x

to the left side

x^{3} - 48 - 12 x = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

x^{3} - 12 x - 48 = 0

Find one solution for

x^{3} - 12 x - 48 = 0

using Newton-Raphson:

x \approx 4.71060 \dots

Apply long division:

\frac{x ^{3} - 12 x - 48}{x - 4.71060 \dots} = x^{2} + 4.71060 \dots x + 10.18977 \dots

x^{2} + 4.71060 \dots x + 10.18977 \dots \approx 0

Find one solution for

x^{2} + 4.71060 \dots x + 10.18977 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

x \approx 4.71060 \dots

x^{3} + 3 x^{2} - 3 x + 1 = 0

245 = 5 x^{3}

x^{3} - 30 x^{2} - 625 x - 3300 = 0

x^{2} \cdot (x + 1)^{2} = 25

2 t^{3} - 2 t = 0

What are the solutions to the equation 12(x+4)=x^3 ?
The solutions to the equation 12(x+4)=x^3 are x\approx 4.71060…
Find the zeros of 12(x+4)=x^3
The zeros of 12(x+4)=x^3 are x\approx 4.71060…