Question

What is the solution set of 2x(x - 1) = 3?

Answer 1

`\displaystyle \\ 2x(x-1) = 3 \\ 2x^2 - 2x =3 \\ 2x^2 - 2x -3=0 \\ \\ x_(12) = (-b \pm √(b^2-4ac) )/(2a) =(2 \pm √(4+24) )/(4) = \\ \\ =(2 \pm √(28) )/(4) =(2 \pm 2√(7) )/(4) = (1 \pm √(7) )/(2) \\ \\ \boxed{x_1 = (1 + √(7) )/(2) } \\ \\ \boxed{x_2= (1 - √(7) )/(2) }`

Answer 2

The distributive property says that:

`a \cdot(b+c) =\acdot b+ a\cdot c`

Given the equation:

`2x(x-1) =3`

Apply the distributive property:

`2x^2-2x=3`

Subtract 3 from both sides we get;

`2x^2-2x-3=0` ....[1]

For the quadratic equation
`ax^2+bx+c =0` where a, b and c are coefficient then the solution is given by:

`x_(1, 2) = (-b \pm√(b^2-4ac))/(2a)`

On comparing general equation with the equation [1] we have;

a = 2 b = -2 and c= -3

then;

`x_(1, 2) = (-(-2) \pm√((-2)^2-4(2)(-3)))/(2(2))`

Simplify:

`x_(1, 2) = (2 \pm√(4+24))/(4)`

`x_(1, 2) = (2 \pm√(28))/(4)`

or

`x_(1, 2) = (2 \pm 2√(7))/(4) = (1\pm √(7) )/(2)`

⇒
`x_(1) = (1 +√(7))/(2)` and
`x_(2) = (1 -√(7))/(2)`

Therefore, the solution for the given equation are:

`x_(1) = (1 +√(7))/(2)` and
`x_(2) = (1 -√(7))/(2)`