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

Question

Mathematics

Tristian Pratt

1 June, 18:32

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

+2

Answers (2)

Know the Answer?

Aliya · Answer 1

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
In interval notation:

(-6, 6)

Answer:

all possible values of b are:

(-6, 6)

Kaelyn Orr · Answer 2

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).