Not sure how to approach this question: "If the solutions of x^2 + bx + 15 = 0 are integers less than 10 but greater than –10, what are the possible values of b?". I've tried a few thing but no luck.
Any thoughts?
I don't really see a way other than trial and error. But, given the restrictions on the solutions, we can restrict our search down to just a few numbers with some clever observations.
First, let's solve the equation for \(x\) in terms of \(b\). By the quadratic formula, \[x=\frac{-b\pm\sqrt{b^2-60}}{2}.\] Next, both solutions for \(x\) must exist
and lie in \((-10,\;10)\), so we must have \[\frac{-b-\sqrt{b^2-60}}{2}>-10\quad\text{and}\quad\frac{-b+\sqrt{b^2-60}}{2}<10\quad\text{and}\quad b^2-60>0.\] Solving those simultaneously gives: \[b\in\left(-\frac{23}{2},\;-\sqrt{60}\right)\cup\left(\sqrt{60},\;\frac{23}{2}\right).\] Now, I claim that if
both solutions for \(x\) are integers, then \(b\)
must also be an integer. Why? Note that we
average of two integer values
must be of the form \(\dfrac{k}{2}\) where \(k\in\mathbb{Z}\). But we know that the average of the two solutions is \(\dfrac{-b}{2}\) because the graph of this quadratic is
symmetrical about the line \(x=\dfrac{-b}{2}\) (or the turning point). So, \(b=-k\in\mathbb{Z}\).
Noting that \(7<\sqrt{60}<8\), we find that if we combine all the restrictions together, we have \[b\in\{-11,-10,-9,-8,\;8,\;9,\;10,\;11\}.\]Using trial and error, we obtain the final result: \[\boxed{b=\pm8}.\]