SOLUTION: Find all the values of n such that x^2 + bx + (b/2)^2 = n has two real roots. Thank you tutors.

 

A quadratic equation with 0 on the right has two
real roots if the discriminant > 0

We find the discriminant which is 

here a=1, b=b, c=

So the discriminant is

 which we set 







Multiply through by 4



Factor out b2





The critical numbers are 0,2 and -2.

[The critical numbers all cause the discriminant to = 0,
which gives the quadratic only one real solution.
So none of the critical numbers are possible values 
for b.

We place the critical numbers on a number line

---------o-------o-------o--------
-4  -3  -2  -1   0   1   2   3   4

We test values in each interval:

Using test value -3 for the first (leftmost) interval







That's true so we shade the interval (-2,0).

---------o=======o-------o--------
-4  -3  -2  -1   0   1   2   3   4

Using test value 1 for the next interval








That's true so we also shade the interval (0,2).

---------o=======o=======o--------
-4  -3  -2  -1   0   1   2   3   4


Using test value 3 for the next interval







That's false.  Therefore b must be in these intervals
in order for the equation to have 2 real roots:

(-2,0) U (0,2)

Edwin