What are the solutions to the equation x(2x+0.25)^2=5.16*10^{-1}1 ?

The solutions to the equation x(2x+0.25)^2=5.16*10^{-1}1 are x\approx 0.42556…

Find the zeros of x(2x+0.25)^2=5.16*10^{-1}1

The zeros of x(2x+0.25)^2=5.16*10^{-1}1 are x\approx 0.42556…

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x (2 x + 0.25)^{2} = 5.16 \cdot 1 0^{- 1} 1

x \approx 0.42556 \dots

Solution steps

x (2 x + 0.25)^{2} = 5.16 \cdot 1 0^{- 1} \cdot 1

Expand

x (2 x + 0.25)^{2} : 4 x^{3} + x^{2} + 0.0625 x

Expand

5.16 \cdot 1 0^{- 1} \cdot 1 : 0.516

4 x^{3} + x^{2} + 0.0625 x = 0.516

Move

0.516

to the left side

4 x^{3} + x^{2} + 0.0625 x - 0.516 = 0

Find one solution for

4 x^{3} + x^{2} + 0.0625 x - 0.516 = 0

using Newton-Raphson:

x \approx 0.42556 \dots

Apply long division:

\frac{4 x ^{3} + x ^{2} + 0.0625 x - 0.516}{x - 0.42556 \dots} = 4 x^{2} + 2.70227 \dots x + 1.21249 \dots

4 x^{2} + 2.70227 \dots x + 1.21249 \dots \approx 0

Find one solution for

4 x^{2} + 2.70227 \dots x + 1.21249 \dots = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

x \approx 0.42556 \dots

c^{6} - 36 c^{4} = 0

x^{4} - 13 x^{2} + 13 = 0

x^{3} - [x] = 4

x^{3} - [x] = 3

x^{4} - 13 x + 36 = 0

What are the solutions to the equation x(2x+0.25)^2=5.16*10^{-1}1 ?
The solutions to the equation x(2x+0.25)^2=5.16*10^{-1}1 are x\approx 0.42556…
Find the zeros of x(2x+0.25)^2=5.16*10^{-1}1
The zeros of x(2x+0.25)^2=5.16*10^{-1}1 are x\approx 0.42556…