What are the solutions to the equation 1-6x(1-x)=0 ?

The solutions to the equation 1-6x(1-x)=0 are x=(3+sqrt(3))/6 ,x=(3-sqrt(3))/6

Find the zeros of 1-6x(1-x)=0

The zeros of 1-6x(1-x)=0 are x=(3+sqrt(3))/6 ,x=(3-sqrt(3))/6

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1 - 6 x (1 - x) = 0

x = \frac{3 + 3}{6}, x = \frac{3 - 3}{6}

Decimal

x = 0.78867 \dots, x = 0.21132 \dots

Solution steps

1 - 6 x (1 - x) = 0

Expand

1 - 6 x (1 - x) : 1 - 6 x + 6 x^{2}

1 - 6 x + 6 x^{2} = 0

Write in the standard form

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

6 x^{2} - 6 x + 1 = 0

Solve with the quadratic formula

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

(- 6)^{2} - 4 \cdot 6 \cdot 1 = 23

x_{1, 2} = \frac{- ( - 6 ) \pm 2 3}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 3}{2 \cdot 6}, x_{2} = \frac{- ( - 6 ) - 2 3}{2 \cdot 6}

x = \frac{- ( - 6 ) + 2 3}{2 \cdot 6} : \frac{3 + 3}{6}

x = \frac{- ( - 6 ) - 2 3}{2 \cdot 6} : \frac{3 - 3}{6}

The solutions to the quadratic equation are:

x = \frac{3 + 3}{6}, x = \frac{3 - 3}{6}

2 k^{2} + 11 k + 12 = 0

8 x^{2} - 10 = 3 x^{2} + 3

\frac{3}{4} (x + 3)^{2} = - 96

y, x^{2} + 4 x y + y^{2} = 0

y, 2 x^{2} = y^{2}

What are the solutions to the equation 1-6x(1-x)=0 ?
The solutions to the equation 1-6x(1-x)=0 are x=(3+sqrt(3))/6 ,x=(3-sqrt(3))/6
Find the zeros of 1-6x(1-x)=0
The zeros of 1-6x(1-x)=0 are x=(3+sqrt(3))/6 ,x=(3-sqrt(3))/6