What are the solutions to the equation 3x^3+2x-1=8x^2-4 ?

The solutions to the equation 3x^3+2x-1=8x^2-4 are x=1,x=(5+sqrt(61))/6 ,x=(5-sqrt(61))/6

Find the zeros of 3x^3+2x-1=8x^2-4

The zeros of 3x^3+2x-1=8x^2-4 are x=1,x=(5+sqrt(61))/6 ,x=(5-sqrt(61))/6

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3 x^{3} + 2 x - 1 = 8 x^{2} - 4

x = 1, x = \frac{5 + 61}{6}, x = \frac{5 - 61}{6}

Decimal Notation

x = 1, x = 2.13504 \dots, x = - 0.46837 \dots

Solution steps

3 x^{3} + 2 x - 1 = 8 x^{2} - 4

Move

4

to the left side

3 x^{3} + 2 x + 3 = 8 x^{2}

Move

8 x^{2}

to the left side

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

Write in the standard form

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

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

Factor

3 x^{3} - 8 x^{2} + 2 x + 3 : (x - 1) (3 x^{2} - 5 x - 3)

(x - 1) (3 x^{2} - 5 x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 3 x^{2} - 5 x - 3 = 0

Solve

x - 1 = 0 : x = 1

Solve

3 x^{2} - 5 x - 3 = 0 : x = \frac{5 + 61}{6}, x = \frac{5 - 61}{6}

The solutions are

x = 1, x = \frac{5 + 61}{6}, x = \frac{5 - 61}{6}

18 x^{3} + 90 x + 55 = 0

x^{4} + x - 4 = 0

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

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

x^{3} - 2 x^{2} - 6 x + 12 = 0

What are the solutions to the equation 3x^3+2x-1=8x^2-4 ?
The solutions to the equation 3x^3+2x-1=8x^2-4 are x=1,x=(5+sqrt(61))/6 ,x=(5-sqrt(61))/6
Find the zeros of 3x^3+2x-1=8x^2-4
The zeros of 3x^3+2x-1=8x^2-4 are x=1,x=(5+sqrt(61))/6 ,x=(5-sqrt(61))/6