Question

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

Answer 1

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

Answer 2

Answer:

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:

$x_{1, 2} = \frac{-b \pm\sqrt{b^2-4ac}}{2a}$

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

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

then;

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

Simplify:

$x_{1, 2} = \frac{2 \pm\sqrt{4+24}}{4}$

$x_{1, 2} = \frac{2 \pm\sqrt{28}}{4}$

or

$x_{1, 2} = \frac{2 \pm 2\sqrt{7}}{4} = \frac{1\pm \sqrt{7} }{2}$

⇒ $x_{1} = \frac{1 +\sqrt{7}}{2}$ and  $x_{2} = \frac{1 -\sqrt{7}}{2}$

Therefore, the solution for the given equation are:

$x_{1} = \frac{1 +\sqrt{7}}{2}$ and  $x_{2} = \frac{1 -\sqrt{7}}{2}$