What are the solutions to the equation a(4a^2+1)= 232/1000 ?

The solutions to the equation a(4a^2+1)= 232/1000 are a= 1/5

Find the zeros of a(4a^2+1)= 232/1000

The zeros of a(4a^2+1)= 232/1000 are a= 1/5

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a (4 a^{2} + 1) = \frac{232}{1000}

a = \frac{1}{5}

Decimal Notation

a = 0.2

Solution steps

a (4 a^{2} + 1) = \frac{232}{1000}

Multiply both sides by

1000

1000 a (4 a^{2} + 1) = 232

Expand

1000 a (4 a^{2} + 1) : 4000 a^{3} + 1000 a

4000 a^{3} + 1000 a = 232

Move

232

to the left side

4000 a^{3} + 1000 a - 232 = 0

Divide both sides by

4000

\frac{4000 a ^{3}}{4000} + \frac{1000 a}{4000} - \frac{232}{4000} = \frac{0}{4000}

Write in the standard form

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

a^{3} + \frac{1 \cdot a}{4} - \frac{29}{500} = 0

Find one solution for

a^{3} + 0.25 a - 0.058 = 0

using Newton-Raphson:

a \approx 0.20000 \dots

Apply long division:

\frac{a ^{3} + \frac{1 \cdot a}{4} - \frac{29}{500}}{a - 0.2} = a^{2} + 0.2 a + 0.29

a^{2} + 0.2 a + 0.29 \approx 0

Find one solution for

a^{2} + 0.2 a + 0.29 = 0

using Newton-Raphson:

No Solution for

a \in R

Approximate to algebraic form

a \approx 0.20000 \dots : a = \frac{1}{5}

a = \frac{1}{5}

The solution is

a = \frac{1}{5}

3 x^{4} - 16 x^{3} + 26 x^{2} - 16 x + 3 = 0

8 x^{4} + 20 x = 20

x^{2} x - 2 = 0

d^{3} + 3 d^{2} + 3 d + 2 = 0

3 x^{3} + 2 x^{2} + x + 1 = 0

What are the solutions to the equation a(4a^2+1)= 232/1000 ?
The solutions to the equation a(4a^2+1)= 232/1000 are a= 1/5
Find the zeros of a(4a^2+1)= 232/1000
The zeros of a(4a^2+1)= 232/1000 are a= 1/5