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

The solutions to the equation (x-3)^2-(x-1)/3 =2x are x=7,x= 4/3

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

The zeros of (x-3)^2-(x-1)/3 =2x are x=7,x= 4/3

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(x - 3)^{2} - \frac{x - 1}{3} = 2 x

x = 7, x = \frac{4}{3}

Solution steps

(x - 3)^{2} - \frac{x - 1}{3} = 2 x

Multiply both sides by

3

3 (x - 3)^{2} - (x - 1) = 6 x

Expand

3 (x - 3)^{2} - (x - 1) : 3 x^{2} - 19 x + 28

3 x^{2} - 19 x + 28 = 6 x

Move

6 x

to the left side

3 x^{2} - 25 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 25 ) \pm ( - 25 ) ^{2} - 4 \cdot 3 \cdot 28}{2 \cdot 3}

(- 25)^{2} - 4 \cdot 3 \cdot 28 = 17

x_{1, 2} = \frac{- ( - 25 ) \pm 17}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 25 ) + 17}{2 \cdot 3}, x_{2} = \frac{- ( - 25 ) - 17}{2 \cdot 3}

x = \frac{- ( - 25 ) + 17}{2 \cdot 3} : 7

x = \frac{- ( - 25 ) - 17}{2 \cdot 3} : \frac{4}{3}

The solutions to the quadratic equation are:

x = 7, x = \frac{4}{3}

0 - \frac{π}{2}

- 5 x^{3} - 75 x^{2} - 130 x

6 f > 12

- 7 - \frac{5}{2}

\frac{a ^{- 1} - b ^{- 1}}{a ^{- 1} + b ^{- 1}}

What are the solutions to the equation (x-3)^2-(x-1)/3 =2x ?
The solutions to the equation (x-3)^2-(x-1)/3 =2x are x=7,x= 4/3
Find the zeros of (x-3)^2-(x-1)/3 =2x
The zeros of (x-3)^2-(x-1)/3 =2x are x=7,x= 4/3