### Instruction: If $x^2+bx+9$ has two non-real roots, find all real possible values of $b$. Express your answer in interval notation.### Response: I know that a quadratic equation of the form $ax^2+bx+c$ has two non-real roots if and only if its discriminant, $b^2-4ac$, is negative.
So I can use this condition to find the values of $b$ that make the discriminant of $x^2+bx+9$ negative.
That is, I want to solve the inequality $b^2-4(1)(9)<0$ for $b$.
I can simplify this inequality by dividing both sides by 4, getting $(b/2)^2-9<0$.
To solve this, I can add 9 to both sides and then take the square root, remembering to include both the positive and negative roots.
This gives me $|b/2|<\sqrt{9}$, or equivalently, $-\sqrt{9}<b/2<\sqrt{9}$.
Multiplying everything by 2, I get $-2\sqrt{9}<b<2\sqrt{9}$.
This is the interval of real values of $b$ that make the quadratic have two non-real roots.