# How do you find the discriminant of x^2-2x+1=0 and use it to determine if the equation has one, two real or two imaginary roots?

May 13, 2017

$\Delta = 0 \to$ equation has a one real solution $x = 1$

#### Explanation:

Consider the general form of the quadratic equation:

$a {x}^{2} + b x + c = 0$

The discriminant $\left(\Delta\right)$ is defined as: ${b}^{2} - 4 a c$

Three cases arise:

(i) $\Delta > 0 \to$ the equation has two distinct real roots
(ii) $\Delta < 0 \to$ the equation has two complex roots
(iii) $\Delta = 0 \to$ the equation has one real solution (Strictly, two equal real roots)

In our equation: ${x}^{2} - 2 x + 1 = 0$

$\Delta = {\left(- 2\right)}^{2} - 4 \cdot 1 \cdot 1 = 4 - 4 = 0$

Hence the equation has one real solution.

The equation may be factorised as: $\left(x - 1\right) \left(x - 1\right) = 0$

Hence the only real solution is $x = 1$