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

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

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

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

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

x = 1, x = \frac{3}{2}

Solution steps

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

Expand

(x - 2) (1 - 2 x) : - 2 x^{2} + 5 x - 2

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

Move

1

to the left side

- 2 x^{2} + 5 x - 3 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1, x = \frac{3}{2}

27 x^{2} + 18 x = 0

24 x^{2} - 2 x - 1 = 0

5 y^{2} + 30 = y^{2} + 6

2 x^{2} - 5 = 9

2 x^{2} - 5 = 8

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