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

The solutions to the equation (2x+1)^4+4(2x+1)^2=12 are x\approx 0.20710…,x\approx-1.20710…

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

The zeros of (2x+1)^4+4(2x+1)^2=12 are x\approx 0.20710…,x\approx-1.20710…

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

x \approx 0.20710 \dots, x \approx - 1.20710 \dots

Solution steps

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

Expand

(2 x + 1)^{4} + 4 (2 x + 1)^{2} : 16 x^{4} + 32 x^{3} + 40 x^{2} + 24 x + 5

16 x^{4} + 32 x^{3} + 40 x^{2} + 24 x + 5 = 12

Move

12

to the left side

16 x^{4} + 32 x^{3} + 40 x^{2} + 24 x - 7 = 0

Find one solution for

16 x^{4} + 32 x^{3} + 40 x^{2} + 24 x - 7 = 0

using Newton-Raphson:

x \approx 0.20710 \dots

Apply long division:

\frac{16 x ^{4} + 32 x ^{3} + 40 x ^{2} + 24 x - 7}{x - 0.20710 \dots} = 16 x^{3} + 35.31370 \dots x^{2} + 47.31370 \dots x + 33.79898 \dots

16 x^{3} + 35.31370 \dots x^{2} + 47.31370 \dots x + 33.79898 \dots \approx 0

Find one solution for

16 x^{3} + 35.31370 \dots x^{2} + 47.31370 \dots x + 33.79898 \dots = 0

using Newton-Raphson:

x \approx - 1.20710 \dots

Apply long division:

\frac{16 x ^{3} + 35.31370 \dots x ^{2} + 47.31370 \dots x + 33.79898 \dots}{x + 1.20710 \dots} = 16 x^{2} + 15.99999 \dots x + 28

16 x^{2} + 15.99999 \dots x + 28 \approx 0

Find one solution for

16 x^{2} + 15.99999 \dots x + 28 = 0

using Newton-Raphson:

No Solution for

x \in R

The solutions are

x \approx 0.20710 \dots, x \approx - 1.20710 \dots

x^{4} - 6 x^{2} + 9 = 0

3 x^{3} + 5 x^{2} + 4 x - 1 = 0

2 x^{3} - 3 x^{2} - 8 x - 3 = 0

v^{3} - 2 v^{2} - 16 v = - 32

r^{4} - 2 r^{3} + 10 r^{2} - 18 r + 9 = 0

What are the solutions to the equation (2x+1)^4+4(2x+1)^2=12 ?
The solutions to the equation (2x+1)^4+4(2x+1)^2=12 are x\approx 0.20710…,x\approx-1.20710…
Find the zeros of (2x+1)^4+4(2x+1)^2=12
The zeros of (2x+1)^4+4(2x+1)^2=12 are x\approx 0.20710…,x\approx-1.20710…