What are the solutions to the equation x^2-(x^3+12)=x-2(x^4-1) ?

The solutions to the equation x^2-(x^3+12)=x-2(x^4-1) are x\approx-1.40668…,x\approx 1.72999…

Find the zeros of x^2-(x^3+12)=x-2(x^4-1)

The zeros of x^2-(x^3+12)=x-2(x^4-1) are x\approx-1.40668…,x\approx 1.72999…

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Popular >

x^2-(x^3+12)=x-2(x^4-1)

Solution

x^{2} - (x^{3} + 12) = x - 2 (x^{4} - 1)

Solution

x \approx - 1.40668 \dots, x \approx 1.72999 \dots

Solution steps

x^{2} - (x^{3} + 12) = x - 2 (x^{4} - 1)

Expand

x^{2} - (x^{3} + 12) : x^{2} - x^{3} - 12

Expand

x - 2 (x^{4} - 1) : x - 2 x^{4} + 2

x^{2} - x^{3} - 12 = x - 2 x^{4} + 2

Switch sides

x - 2 x^{4} + 2 = x^{2} - x^{3} - 12

Move

12

to the left side

- 2 x^{4} + x + 14 = x^{2} - x^{3}

Move

x^{3}

to the left side

- 2 x^{4} + x + 14 + x^{3} = x^{2}

Move

x^{2}

to the left side

- 2 x^{4} + x + 14 + x^{3} - x^{2} = 0

Write in the standard form

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

- 2 x^{4} + x^{3} - x^{2} + x + 14 = 0

Find one solution for

- 2 x^{4} + x^{3} - x^{2} + x + 14 = 0

using Newton-Raphson:

x \approx - 1.40668 \dots

Apply long division:

\frac{- 2 x ^{4} + x ^{3} - x ^{2} + x + 14}{x + 1.40668 \dots} = - 2 x^{3} + 3.81337 \dots x^{2} - 6.36422 \dots x + 9.95246 \dots

- 2 x^{3} + 3.81337 \dots x^{2} - 6.36422 \dots x + 9.95246 \dots \approx 0

Find one solution for

- 2 x^{3} + 3.81337 \dots x^{2} - 6.36422 \dots x + 9.95246 \dots = 0

using Newton-Raphson:

x \approx 1.72999 \dots

Apply long division:

\frac{- 2 x ^{3} + 3.81337 \dots x ^{2} - 6.36422 \dots x + 9.95246 \dots}{x - 1.72999 \dots} = - 2 x^{2} + 0.35337 \dots x - 5.75287 \dots

- 2 x^{2} + 0.35337 \dots x - 5.75287 \dots \approx 0

Find one solution for

- 2 x^{2} + 0.35337 \dots x - 5.75287 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solutions are

x \approx - 1.40668 \dots, x \approx 1.72999 \dots

Popular Examples

(x-4)(x^2+10x+52)=(x-4)^2

(x - 4) (x^{2} + 10 x + 52) = (x - 4)^{2}

8t^3-72t^2+228t-252=0

8 t^{3} - 72 t^{2} + 228 t - 252 = 0

-19.6x^3+26.067x^2-7.29=0

- 19.6 x^{3} + 26.067 x^{2} - 7.29 = 0

x^3=a

x^{3} = a

t^4+3t^3-10t^2=0

t^{4} + 3 t^{3} - 10 t^{2} = 0

Frequently Asked Questions (FAQ)

What are the solutions to the equation x^2-(x^3+12)=x-2(x^4-1) ?
The solutions to the equation x^2-(x^3+12)=x-2(x^4-1) are x\approx-1.40668…,x\approx 1.72999…
Find the zeros of x^2-(x^3+12)=x-2(x^4-1)
The zeros of x^2-(x^3+12)=x-2(x^4-1) are x\approx-1.40668…,x\approx 1.72999…