What are the solutions to the equation 1-3x(1-x)=0 ?

The solutions to the equation 1-3x(1-x)=0 are x= 1/2+i(sqrt(3))/6 ,x= 1/2-i(sqrt(3))/6

Find the zeros of 1-3x(1-x)=0

The zeros of 1-3x(1-x)=0 are x= 1/2+i(sqrt(3))/6 ,x= 1/2-i(sqrt(3))/6

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1 - 3 x (1 - x) = 0

x = \frac{1}{2} + i \frac{3}{6}, x = \frac{1}{2} - i \frac{3}{6}

Solution steps

1 - 3 x (1 - x) = 0

Expand

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 3 \cdot 1 : 3 i

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

Separate the solutions

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

x = \frac{- ( - 3 ) + 3 i}{2 \cdot 3} : \frac{1}{2} + i \frac{3}{6}

x = \frac{- ( - 3 ) - 3 i}{2 \cdot 3} : \frac{1}{2} - i \frac{3}{6}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{3}{6}, x = \frac{1}{2} - i \frac{3}{6}

∣ x - 1 ∣ < 6

2 (x - 7) + 3 x = 4 x - 8 - x

0 = x^{2} + 2 (0.7) x - 0.4125

10 x^{2} + 33 x - 7

\frac{( 3 ab c ) ^{3}}{( 2 a ^{- 1} b ^{- 2} c ) ^{2}}

What are the solutions to the equation 1-3x(1-x)=0 ?
The solutions to the equation 1-3x(1-x)=0 are x= 1/2+i(sqrt(3))/6 ,x= 1/2-i(sqrt(3))/6
Find the zeros of 1-3x(1-x)=0
The zeros of 1-3x(1-x)=0 are x= 1/2+i(sqrt(3))/6 ,x= 1/2-i(sqrt(3))/6