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

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

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

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

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4 x - x^{2} = 1

x = 2 - 3, x = 2 + 3

Decimal Notation

x = 0.26794 \dots, x = 3.73205 \dots

Solution steps

4 x - x^{2} = 1

Move

1

to the left side

4 x - x^{2} - 1 = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

4^{2} - 4 (- 1) (- 1) = 23

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

Separate the solutions

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

x = \frac{- 4 + 2 3}{2 ( - 1 )} : 2 - 3

x = \frac{- 4 - 2 3}{2 ( - 1 )} : 2 + 3

The solutions to the quadratic equation are:

x = 2 - 3, x = 2 + 3

3 x^{2} + 10 x - 4 = 0

- x^{2} - x + 42 = 0

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

- 0.001 x^{2} - 0.4 x + 210 = 0.0006 x^{2} + 0.26 x + 128

(x + 3)^{2} - 9 = 7

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