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

The solutions to the equation (2x^2+1)x^3=12 are x\approx 1.36448…

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

The zeros of (2x^2+1)x^3=12 are x\approx 1.36448…

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

x \approx 1.36448 \dots

Solution steps

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

Expand

(2 x^{2} + 1) x^{3} : 2 x^{5} + x^{3}

2 x^{5} + x^{3} = 12

Move

12

to the left side

2 x^{5} + x^{3} - 12 = 0

Find one solution for

2 x^{5} + x^{3} - 12 = 0

using Newton-Raphson:

x \approx 1.36448 \dots

Apply long division:

\frac{2 x ^{5} + x ^{3} - 12}{x - 1.36448 \dots} = 2 x^{4} + 2.72896 \dots x^{3} + 4.72363 \dots x^{2} + 6.44531 \dots x + 8.79453 \dots

2 x^{4} + 2.72896 \dots x^{3} + 4.72363 \dots x^{2} + 6.44531 \dots x + 8.79453 \dots \approx 0

Find one solution for

2 x^{4} + 2.72896 \dots x^{3} + 4.72363 \dots x^{2} + 6.44531 \dots x + 8.79453 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

x \approx 1.36448 \dots

(z^{2} + 6)^{4} = 0

x^{3} - 4 x^{2} + 7 x - 4 = 0

x^{3} + 2 x^{2} - x + 6 = 0

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

x^{1} - 2 x^{2} + 3 x^{3} = 11

What are the solutions to the equation (2x^2+1)x^3=12 ?
The solutions to the equation (2x^2+1)x^3=12 are x\approx 1.36448…
Find the zeros of (2x^2+1)x^3=12
The zeros of (2x^2+1)x^3=12 are x\approx 1.36448…