Algeb problem need help about roots?

Question

23. What are the roots of the equation (2x-1)(x-1) = 3 ?

Answer 1

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

First multiply out the factors.  Then bring the constant over to the left side of the equation so that you can set the expression equal to 0.

2x² - 3x + 1 = 3

2x² - 3x + 1 - 3 = 3 - 3

2x² - 3x - 2 = 0.

Now factor the left side of the equation:

(2x + 1)(x - 2) = 0.

Now set each of the factors equal to 0, and solve for x in each factor:

2x + 1 = 0

2x = -1

x = -½

x - 2 = 0

x = 2

So x = -½ and x = 2 seem to be the roots of the equation.  To verify they are correct, substitute both of them into the original equation.

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

[2 (-½) - 1](-½ - 1) = 3

(-1 - 1)(-3/2) = 3

(-2)(-3/2) = 3

(6/2) = 3

3 = 3

x = -½ works.

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

[2 (2) - 1](2 - 1) = 3

(4 - 1)(2 - 1) = 3

3 (1) = 3

3 = 3

x = 2 also works.  So we have the correct solutions.  x = -½ or x = 2 are the roots of the equation.

Answer 2

2x²-2x-x+1=3

2x²-3x-2=0

(2x+1)(x-2)=0

x=-1/2 or x=2

So the roots are -1/2 and 2.

Answer 3

-6 and 5 that is if u took into account for x that gets delivered to its secondary