If $x^2+bx+9$ has two non-real roots, find all possible values of $b$. express your answer in interval notation.

In order to have non-real roots, the radicand must be less than zero, therefore:
x² + bx + 9 = 0
will have as a radicand b² - 4·1·9

Therefore:
b² - 36 < 0

First, find the solutions to the equation b² - 36 = 0
which are b = +/- 6

Since the disequality asks for "less than", we need to take the inner interval between these two solutions:
-6 < b < +6
or else, ] -6 ; +6 [
or else, ( -6 ; +6 )
(These are all valid notations).

Answer Link

carlosego carlosego · Answer 2 · 2018-04-04T13:05:42+02:00

carlosego carlosego

04-04-2018

Applying the resolver we have:
x = (- b +/- Root (b ^ 2 - 4 * a * c)) / (2 * a)
Substituting values we have:
x = (- b +/- Root (b ^ 2 - 4 * 1 * 9)) / (2 * 1)
Rewriting:
x = (- b +/- Root (b ^ 2 - 36)) / (2)
For non-real values we have:
b ^ 2 - 36 <0
We cleared b:
b ^ 2 <36
b <+/- root (36)
b <+/- 6
Rewriting:
-6 <b <6
In interval notation:
(-6, 6)
Answer:
all possible values of b are:
(-6, 6)

Answer Link