# How do you find the nature of the roots using the discriminant given x^2 - 7x + 12 = 0?

Sep 25, 2016

Roots are rational. These are $x = 3$ or $x = 4$.

#### Explanation:

If the equation is $a {x}^{2} + b x + c = 0$, nature of roots is decided by discriminant ${b}^{2} - 4 a c$

If $a$, $b$ and $c$ are rational and ${b}^{2} - 4 a c$ is square of a rational number, roots are rational.

If ${b}^{2} - 4 a c > 0$ but is not a square of a rational number, roots are real but not rational.

If ${b}^{2} - 4 a c > 0 - 0$ we have equal roots.

If ${b}^{2} - 4 a c < 0$ roots are complex

In ${x}^{2} - 7 x + 12 = 0$, discriminant is ${\left(- 7\right)}^{2} - 4 \times 1 \times 12 = 49 - 48 = 1 = {1}^{2}$

hence roots are rational. In fact

${x}^{2} - 7 x + 12 = 0$

$\Leftrightarrow {x}^{2} - 4 x - 3 x + 12 = 0$

or $x \left(x - 4\right) - 3 \left(x - 4\right) = 0$

or $\left(x - 3\right) \left(x - 4\right) = 0$ i.e. $x = 3$ or $x = 4$