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

The solutions to the equation 3= 1/4 x^3+x-1 are x=2,x=-1+sqrt(7)i,x=-1-sqrt(7)i

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

The zeros of 3= 1/4 x^3+x-1 are x=2,x=-1+sqrt(7)i,x=-1-sqrt(7)i

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3 = \frac{1}{4} x^{3} + x - 1

x = 2, x = - 1 + 7 i, x = - 1 - 7 i

Solution steps

3 = \frac{1}{4} x^{3} + x - 1

Multiply both sides by

4

12 = x^{3} + 4 x - 4

Switch sides

x^{3} + 4 x - 4 = 12

Move

12

to the left side

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

Factor

x^{3} + 4 x - 16 : (x - 2) (x^{2} + 2 x + 8)

(x - 2) (x^{2} + 2 x + 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

x^{2} + 2 x + 8 = 0 : x = - 1 + 7 i, x = - 1 - 7 i

The solutions are

x = 2, x = - 1 + 7 i, x = - 1 - 7 i

s^{3} + s^{2} + 4 s + 4 = 0

(- 11 x^{2} + 62) (- 9 x^{2} + 25) - (- 11) (- 11) = 0

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

10 x^{3} - 39 x^{2} + 29 x - 6 = 0

4 x^{3} - 54 x^{2} + 240 x - 350 = 0

What are the solutions to the equation 3= 1/4 x^3+x-1 ?
The solutions to the equation 3= 1/4 x^3+x-1 are x=2,x=-1+sqrt(7)i,x=-1-sqrt(7)i
Find the zeros of 3= 1/4 x^3+x-1
The zeros of 3= 1/4 x^3+x-1 are x=2,x=-1+sqrt(7)i,x=-1-sqrt(7)i