How do you find the value of the discriminant and state the type of solutions given `r^2+5r+2=0`?

The discriminant of a quadratic equation can be calculated as:

`Delta=b^2-4*a*c`

Here we have:

`a=1`, `b=5`, `c=2`

So the discriminant is:

`Delta=5^2-4*1*2=25-8=17`

To state the type and number of solutions we use the following property of a quadratic equation:

The quadratic equation has:

• no real solutions if `Delta<0`
• one real solution if `Delta=0`
• two real solutions if `Delta>0`

Here the discriminant is positive, so the equation has 2 different real solutions.

Given:

`r^2+5r+2 = 0`

Note that this is of the form:

`ar^2+br+c = 0`

with `a=1`, `b=5` and `c=2`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac`

`color(white)(Delta) = color(blue)(5)^2-4(color(blue)(1))(color(blue)(2))`

`color(white)(Delta)= 25-8`

`color(white)(Delta)= 17`

Since `Delta > 0` this quadratic has two distinct real roots, but because `Delta` is not a perfect square those roots are irrational.

The possible cases are:

• `Delta > 0` with `Delta` a perfect square: Two distinct rational roots.

• `Delta > 0` with `Delta` not a perfect square: Two distinct irrational roots.

• `Delta = 0`: One repeated rational root.

• `Delta < 0`: Two distinct non-real Complex roots, which are complex conjugates of one another.

`color(white)()`
Bonus

We can find the roots using the quadratic formula:

`r = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(r) = (-b+-sqrt(Delta))/(2a)`

`color(white)(r) = (-5+-sqrt(17))/2`

`color(white)(r) = -5/2+-sqrt(17)/2`