What are the solutions to the equation 17c^1+5c^2+255c^3=53 ?

The solutions to the equation 17c^1+5c^2+255c^3=53 are c\approx 0.54886…

Find the zeros of 17c^1+5c^2+255c^3=53

The zeros of 17c^1+5c^2+255c^3=53 are c\approx 0.54886…

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17 c^{1} + 5 c^{2} + 255 c^{3} = 53

c \approx 0.54886 \dots

Solution steps

17 c^{1} + 5 c^{2} + 255 c^{3} = 53

Refine

17 c + 5 c^{2} + 255 c^{3} = 53

Move

53

to the left side

17 c + 5 c^{2} + 255 c^{3} - 53 = 0

Write in the standard form

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

255 c^{3} + 5 c^{2} + 17 c - 53 = 0

Find one solution for

255 c^{3} + 5 c^{2} + 17 c - 53 = 0

using Newton-Raphson:

c \approx 0.54886 \dots

Apply long division:

\frac{255 c ^{3} + 5 c ^{2} + 17 c - 53}{c - 0.54886 \dots} = 255 c^{2} + 144.96008 \dots c + 96.56324 \dots

255 c^{2} + 144.96008 \dots c + 96.56324 \dots \approx 0

Find one solution for

255 c^{2} + 144.96008 \dots c + 96.56324 \dots = 0

using Newton-Raphson:

No Solution for

c \in R

The solution is

c \approx 0.54886 \dots

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

(2 x + 2) (3 x - 6) (4 x + 12) = 0

(x^{2} + 2) (x^{2} - 12) = 72

- 5 + 4 m^{3} = 103

- 6 x^{4} + 8 x^{2} + 2 = 0

What are the solutions to the equation 17c^1+5c^2+255c^3=53 ?
The solutions to the equation 17c^1+5c^2+255c^3=53 are c\approx 0.54886…
Find the zeros of 17c^1+5c^2+255c^3=53
The zeros of 17c^1+5c^2+255c^3=53 are c\approx 0.54886…