What are the solutions to the equation 4x-x^2=x+1 ?

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

Find the zeros of 4x-x^2=x+1

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

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

x = \frac{3 - 5}{2}, x = \frac{3 + 5}{2}

Decimal Notation

x = 0.38196 \dots, x = 2.61803 \dots

Solution steps

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

Move

1

to the left side

4 x - x^{2} - 1 = x

Move

x

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

3^{2} - 4 (- 1) (- 1) = 5

x_{1, 2} = \frac{- 3 \pm 5}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 3 + 5}{2 ( - 1 )}, x_{2} = \frac{- 3 - 5}{2 ( - 1 )}

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

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

The solutions to the quadratic equation are:

x = \frac{3 - 5}{2}, x = \frac{3 + 5}{2}

m^{2} - \frac{49}{2} m = \frac{1}{2}

a, y = (a x + b)^{2}

1 2^{2} + b^{2} = 3 7^{2}

12 + 1.6 x = 8 x - \frac{1}{5} x^{2}

(3 - x) (- 1 - x) + 4 = 0

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