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

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

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

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

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

x = 1, x = \frac{1}{4}

Decimal Notation

x = 1, x = 0.25

Solution steps

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

Expand

x^{2} + (1 - x) (1 - 3 x) : 4 x^{2} - 4 x + 1

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

Move

x

to the left side

4 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

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

x_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 4}, x_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 4}

x = \frac{- ( - 5 ) + 3}{2 \cdot 4} : 1

x = \frac{- ( - 5 ) - 3}{2 \cdot 4} : \frac{1}{4}

The solutions to the quadratic equation are:

x = 1, x = \frac{1}{4}

a (a - 5) < 2 (a - 5)

\frac{2}{6}

\frac{x - 7}{2 x + 9} \geq 1

x^{2} - 8 x + 64 = 0

(\frac{1}{36})^{x + 8} \leq 21 6^{x - 3}

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