# How do you use the discriminant to determine the nature of the solutions given  x^2 – 6x + 5 = 0?

Apr 27, 2017

#### Explanation:

For an equation $a {x}^{2} + b x + c = 0$, (assume $a , b$ and $c$ are rational numbers), the discriminant is ${b}^{2} - 4 a c$.

If ${b}^{2} - 4 a c = 0$, we have two roots which are equal i.e. repeated roots.

if ${b}^{2} - 4 a c > 0$ and is square of a rational number, then it has two distinct roots, which are rational.

if ${b}^{2} - 4 a c > 0$ and but is not square of a rational number, then the two roots are irrational.

if ${b}^{2} - 4 a c < 0$, then the two roots are complex conjugate numbers.

Here, in ${x}^{2} - 6 x + 5 = 0$, we have discriminant ${\left(- 6\right)}^{2} - 4 \times 1 \times 5 = 16$, which is square of rational number $2$,

hence roots are two distinct rational numbers.

In fact as ${x}^{2} - 6 x + 5 = 0 \Leftrightarrow \left(x - 1\right) \left(x - 5\right) - 0$, roots are $1$ and $5$.