What are the solutions to the equation 3/1+x-x^2+1/1-x^2+x+1/1-x=0 ?

The solutions to the equation 3/1+x-x^2+1/1-x^2+x+1/1-x=0 are x=-(-1+sqrt(41))/4 ,x=(1+sqrt(41))/4

Find the zeros of 3/1+x-x^2+1/1-x^2+x+1/1-x=0

The zeros of 3/1+x-x^2+1/1-x^2+x+1/1-x=0 are x=-(-1+sqrt(41))/4 ,x=(1+sqrt(41))/4

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

x = - \frac{- 1 + 41}{4}, x = \frac{1 + 41}{4}

Decimal Notation

x = - 1.35078 \dots, x = 1.85078 \dots

Solution steps

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

Expand

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 5}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 5 = 41

x_{1, 2} = \frac{- 1 \pm 41}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 1 + 41}{2 ( - 2 )}, x_{2} = \frac{- 1 - 41}{2 ( - 2 )}

x = \frac{- 1 + 41}{2 ( - 2 )} : - \frac{- 1 + 41}{4}

x = \frac{- 1 - 41}{2 ( - 2 )} : \frac{1 + 41}{4}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 41}{4}, x = \frac{1 + 41}{4}

\frac{x}{3} = \frac{x ^{2}}{5}

x^{2} + 2 a - 15 = 0

9 x^{2} - 60 x + (a^{2} + b^{2}) = 0

x, y^{4} = c \cdot x^{2}

e = \frac{h ^{2} k ^{2}}{2 m}

What are the solutions to the equation 3/1+x-x^2+1/1-x^2+x+1/1-x=0 ?
The solutions to the equation 3/1+x-x^2+1/1-x^2+x+1/1-x=0 are x=-(-1+sqrt(41))/4 ,x=(1+sqrt(41))/4
Find the zeros of 3/1+x-x^2+1/1-x^2+x+1/1-x=0
The zeros of 3/1+x-x^2+1/1-x^2+x+1/1-x=0 are x=-(-1+sqrt(41))/4 ,x=(1+sqrt(41))/4