# How do you use the discriminant to determine the nature of the roots for 5x^2 - 5x – 60 = 0?

Jun 15, 2015

$5 {x}^{2} - 5 x - 60 = 5 \left({x}^{2} - x - 12\right)$

$\Delta \left({x}^{2} - x - 12\right) = {7}^{2}$

Since this is positive and a perfect square, the two roots are distinct, real and rational.

#### Explanation:

$5 {x}^{2} - 5 x - 60 = 5 \left({x}^{2} - x - 12\right)$

${x}^{2} - x - 12$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 1$ and $c = - 12$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - \left(4 \times 1 \times - 12\right) = 1 + 48 = 49 = {7}^{2}$

Since this is positive and a perfect square, ${x}^{2} - x - 12 = 0$
and hence $5 {x}^{2} - 5 x - 60 = 0$ has two distinct real, rational roots.

Here are the possible cases:

$\Delta > 0$ There are two distinct, real roots. If $\Delta$ is also a perfect square (and the original coefficients are rational), then the roots are also rational.

$\Delta = 0$ There is one repeated root (with multiplicity 2). If the coefficients of the quadratic are rational, this root is rational too.

$\Delta < 0$ There are no real roots. There are two distinct complex roots (which are complex conjugates of one another).