What are the solutions to the equation x^2-4=-x^3+6 ?

The solutions to the equation x^2-4=-x^3+6 are x\approx 1.86746…

Find the zeros of x^2-4=-x^3+6

The zeros of x^2-4=-x^3+6 are x\approx 1.86746…

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x^{2} - 4 = - x^{3} + 6

x \approx 1.86746 \dots

Solution steps

x^{2} - 4 = - x^{3} + 6

Switch sides

- x^{3} + 6 = x^{2} - 4

Move

4

to the left side

- x^{3} + 10 = x^{2}

Move

x^{2}

to the left side

- x^{3} + 10 - x^{2} = 0

Write in the standard form

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

- x^{3} - x^{2} + 10 = 0

Find one solution for

- x^{3} - x^{2} + 10 = 0

using Newton-Raphson:

x \approx 1.86746 \dots

Apply long division:

\frac{- x ^{3} - x ^{2} + 10}{x - 1.86746 \dots} = - x^{2} - 2.86746 \dots x - 5.35486 \dots

- x^{2} - 2.86746 \dots x - 5.35486 \dots \approx 0

Find one solution for

- x^{2} - 2.86746 \dots x - 5.35486 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

x \approx 1.86746 \dots

0.11 = (1 + \frac{r}{12})^{12} - 1

(- 1 - x)^{2} (- 1 - 2 x - x^{2})^{2} = 0

x (x^{2} - 9) = 0

x^{3} (22 - (\frac{x}{2})) = 18

4 x (x - 1)^{2} = 1

What are the solutions to the equation x^2-4=-x^3+6 ?
The solutions to the equation x^2-4=-x^3+6 are x\approx 1.86746…
Find the zeros of x^2-4=-x^3+6
The zeros of x^2-4=-x^3+6 are x\approx 1.86746…