What are the solutions to the equation x(1-3x)=2x-1 ?

The solutions to the equation x(1-3x)=2x-1 are x=-(1+sqrt(13))/6 ,x=(sqrt(13)-1)/6

Find the zeros of x(1-3x)=2x-1

The zeros of x(1-3x)=2x-1 are x=-(1+sqrt(13))/6 ,x=(sqrt(13)-1)/6

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x (1 - 3 x) = 2 x - 1

x = - \frac{1 + 13}{6}, x = \frac{13 - 1}{6}

Decimal Notation

x = - 0.76759 \dots, x = 0.43425 \dots

Solution steps

x (1 - 3 x) = 2 x - 1

Expand

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

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

Move

1

to the left side

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

Move

2 x

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 3) \cdot 1 = 13

x_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 13}{2 ( - 3 )}, x_{2} = \frac{- ( - 1 ) - 13}{2 ( - 3 )}

x = \frac{- ( - 1 ) + 13}{2 ( - 3 )} : - \frac{1 + 13}{6}

x = \frac{- ( - 1 ) - 13}{2 ( - 3 )} : \frac{13 - 1}{6}

The solutions to the quadratic equation are:

x = - \frac{1 + 13}{6}, x = \frac{13 - 1}{6}

2 x^{2} - 9 x - 13 = - 8

10 x^{2} + 15 = 0

m^{2} + 2 m - 8 = 0

x^{2} - 7 x = - 6

9 x^{2} + 38 = 36 x

What are the solutions to the equation x(1-3x)=2x-1 ?
The solutions to the equation x(1-3x)=2x-1 are x=-(1+sqrt(13))/6 ,x=(sqrt(13)-1)/6
Find the zeros of x(1-3x)=2x-1
The zeros of x(1-3x)=2x-1 are x=-(1+sqrt(13))/6 ,x=(sqrt(13)-1)/6