What are the solutions to the equation 7x^3+2x^2-x=0 ?

The solutions to the equation 7x^3+2x^2-x=0 are x=0,x=(-1+2sqrt(2))/7 ,x=-(1+2sqrt(2))/7

Find the zeros of 7x^3+2x^2-x=0

The zeros of 7x^3+2x^2-x=0 are x=0,x=(-1+2sqrt(2))/7 ,x=-(1+2sqrt(2))/7

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

x = 0, x = \frac{- 1 + 2 2}{7}, x = - \frac{1 + 2 2}{7}

Decimal Notation

x = 0, x = 0.26120 \dots, x = - 0.54691 \dots

Solution steps

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

Factor

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

x (7 x^{2} + 2 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 7 x^{2} + 2 x - 1 = 0

Solve

7 x^{2} + 2 x - 1 = 0 : x = \frac{- 1 + 2 2}{7}, x = - \frac{1 + 2 2}{7}

The solutions are

x = 0, x = \frac{- 1 + 2 2}{7}, x = - \frac{1 + 2 2}{7}

x^{3} - 4 x^{2} - 17 x + 60 = 0

(x^{2} + 1) (x^{2} + 4) = 0

n^{3} - 2 n^{2} + 15 n - 500 = 0

1 x^{4} + 21 x^{2} - 100 = 0

t^{3} = 216

What are the solutions to the equation 7x^3+2x^2-x=0 ?
The solutions to the equation 7x^3+2x^2-x=0 are x=0,x=(-1+2sqrt(2))/7 ,x=-(1+2sqrt(2))/7
Find the zeros of 7x^3+2x^2-x=0
The zeros of 7x^3+2x^2-x=0 are x=0,x=(-1+2sqrt(2))/7 ,x=-(1+2sqrt(2))/7