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

2 Answers
Apr 27, 2017

Answer:

See explanation.

Explanation:

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.

Apr 27, 2017

Answer:

#Delta = 17# telling us that this quadratic equation has two distinct irrational roots.

Explanation:

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#