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 \(x_{2} = \frac{1 -\sqrt{7}}{2}\)
