What are the solutions to the equation 15x^3-x+7=x^3+49x^2+x ?

The solutions to the equation 15x^3-x+7=x^3+49x^2+x are x= 7/2 ,x=-1/(sqrt(7)),x= 1/(sqrt(7))

Find the zeros of 15x^3-x+7=x^3+49x^2+x

The zeros of 15x^3-x+7=x^3+49x^2+x are x= 7/2 ,x=-1/(sqrt(7)),x= 1/(sqrt(7))

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15 x^{3} - x + 7 = x^{3} + 49 x^{2} + x

x = \frac{7}{2}, x = - \frac{1}{7}, x = \frac{1}{7}

Decimal Notation

x = 3.5, x = - 0.37796 \dots, x = 0.37796 \dots

Solution steps

15 x^{3} - x + 7 = x^{3} + 49 x^{2} + x

Move

x

to the left side

15 x^{3} - 2 x + 7 = x^{3} + 49 x^{2}

Move

49 x^{2}

to the left side

15 x^{3} - 2 x + 7 - 49 x^{2} = x^{3}

Move

x^{3}

to the left side

14 x^{3} - 49 x^{2} - 2 x + 7 = 0

Factor

14 x^{3} - 49 x^{2} - 2 x + 7 : (2 x - 7) (7 x + 1) (7 x - 1)

(2 x - 7) (7 x + 1) (7 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x - 7 = 0 or 7 x + 1 = 0 or 7 x - 1 = 0

Solve

2 x - 7 = 0 : x = \frac{7}{2}

Solve

7 x + 1 = 0 : x = - \frac{1}{7}

Solve

7 x - 1 = 0 : x = \frac{1}{7}

The solutions are

x = \frac{7}{2}, x = - \frac{1}{7}, x = \frac{1}{7}

x^{3} - 3 x^{2} - 4 x + 12 = 0

x^{3} = 27

x^{3} = - 8

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

x, x^{3} + y^{3} = z^{3}

What are the solutions to the equation 15x^3-x+7=x^3+49x^2+x ?
The solutions to the equation 15x^3-x+7=x^3+49x^2+x are x= 7/2 ,x=-1/(sqrt(7)),x= 1/(sqrt(7))
Find the zeros of 15x^3-x+7=x^3+49x^2+x
The zeros of 15x^3-x+7=x^3+49x^2+x are x= 7/2 ,x=-1/(sqrt(7)),x= 1/(sqrt(7))