What are the solutions to the equation 5(-1)^2=x(-1)^3-1+x^2 ?

The solutions to the equation 5(-1)^2=x(-1)^3-1+x^2 are x=3,x=-2

Find the zeros of 5(-1)^2=x(-1)^3-1+x^2

The zeros of 5(-1)^2=x(-1)^3-1+x^2 are x=3,x=-2

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

x = 3, x = - 2

Solution steps

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

Expand

5 (- 1)^{2} : 5

Expand

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

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

Switch sides

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

Move

5

to the left side

x^{2} - x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 6) = 5

x_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

x = \frac{- ( - 1 ) + 5}{2 \cdot 1} : 3

x = \frac{- ( - 1 ) - 5}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 3, x = - 2

4 x^{2} - 6 x - 2 = 0

2 x^{2} - 4 x + 4 = 2

x, x^{2} + x + 6 = 0

\frac{2}{3} a^{2} - 51 = - 3

n (n - 1) = 72

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