How do you find the value of the discriminant and state the type of solutions given #r^2+5r+2=0#?
2 Answers
Answer:
See explanation.
Explanation:
The discriminant of a quadratic equation can be calculated as:
#Delta=b^24*a*c#
Here we have:
So the discriminant is:
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.
Answer:
Explanation:
Given:
#r^2+5r+2 = 0#
Note that this is of the form:
#ar^2+br+c = 0#
with
This has discriminant
#Delta = b^24ac#
#color(white)(Delta) = color(blue)(5)^24(color(blue)(1))(color(blue)(2))#
#color(white)(Delta)= 258#
#color(white)(Delta)= 17#
Since
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 nonreal Complex roots, which are complex conjugates of one another.
Bonus
We can find the roots using the quadratic formula:
#r = (b+sqrt(b^24ac))/(2a)#
#color(white)(r) = (b+sqrt(Delta))/(2a)#
#color(white)(r) = (5+sqrt(17))/2#
#color(white)(r) = 5/2+sqrt(17)/2#