What are the solutions to the equation x^3+4x=5 ?

The solutions to the equation x^3+4x=5 are x=1,x=-1/2+i(sqrt(19))/2 ,x=-1/2-i(sqrt(19))/2

Find the zeros of x^3+4x=5

The zeros of x^3+4x=5 are x=1,x=-1/2+i(sqrt(19))/2 ,x=-1/2-i(sqrt(19))/2

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x^{3} + 4 x = 5

x = 1, x = - \frac{1}{2} + i \frac{19}{2}, x = - \frac{1}{2} - i \frac{19}{2}

Solution steps

x^{3} + 4 x = 5

Move

5

to the left side

x^{3} + 4 x - 5 = 0

Factor

x^{3} + 4 x - 5 : (x - 1) (x^{2} + x + 5)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

x^{2} + x + 5 = 0 : x = - \frac{1}{2} + i \frac{19}{2}, x = - \frac{1}{2} - i \frac{19}{2}

The solutions are

x = 1, x = - \frac{1}{2} + i \frac{19}{2}, x = - \frac{1}{2} - i \frac{19}{2}

x^{3} + 4 x = 8

3 x^{4} - 74 x^{2} - 25 = 0

114 s^{5} = 1.34 \cdot 1 0^{- 34}

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

- 50 = 10 b^{10} b

What are the solutions to the equation x^3+4x=5 ?
The solutions to the equation x^3+4x=5 are x=1,x=-1/2+i(sqrt(19))/2 ,x=-1/2-i(sqrt(19))/2
Find the zeros of x^3+4x=5
The zeros of x^3+4x=5 are x=1,x=-1/2+i(sqrt(19))/2 ,x=-1/2-i(sqrt(19))/2