TMUA practice: Quadratics, problem 3

The discriminant of $\;ax^{2} + 2bx + c = 0\;$ is $\;(2b)^{2} - 4ac = 4(b^{2} - ac)$. ‘No real roots’ means $\;b^{2} - ac < 0$, i.e. $\;b^{2} < ac$ (and $a,\, c > 0$ make $ac > 0$).

Test each progression.

• AP. $\;a,\, b,\, c\;$ in AP means $\;2b = a + c$, so $\;b = (a + c)/2$. Then $\;b^{2} = (a + c)^{2}/4$. The condition $b^{2} < ac$ becomes $\;(a + c)^{2} < 4ac$, i.e. $\;(a - c)^{2} < 0$. Impossible — the LHS is a square. So AP is not possible.
• GP. $\;a,\, b,\, c\;$ in GP means $\;b^{2} = ac\;$ (the standard GP middle-term identity). Then the strict inequality $b^{2} < ac$ becomes $\;ac < ac$, impossible. So GP is not possible.
• HP. $\;a,\, b,\, c\;$ in HP means $\;b = 2ac/(a + c)\;$ (the harmonic-mean formula). Then $\;b^{2} = 4 a^{2} c^{2}/(a + c)^{2}$. The condition $b^{2} < ac$ becomes $\;4 a^{2} c^{2}/(a + c)^{2} < ac$, i.e. (cancelling $ac > 0$) $\;4 a c < (a + c)^{2}$, i.e. $\;(a - c)^{2} > 0$. Possible whenever $a \ne c$.

So $a,\, b,\, c$ cannot be in AP or GP, but can be in HP (with $a \ne c$). This is option C.

Why the other options fail.

• A is exactly backwards on GP and AP: GP requires $b^{2} = ac$, contradicting $b^{2} < ac$ strictly.
• B: the AP case forces $(a - c)^{2} < 0$, impossible.
• D: the HP case is consistent (e.g. $a = 1,\, c = 4$ in HP gives $b = 8/5$ and $b^{2} = 64/25 = 2.56 < 4 = ac$).

The lesson: AM-GM-HM inequality states $\text{AM} \ge \text{GM} \ge \text{HM}$. AP/GP/HP impose $b = \text{AM}$, $b^{2} = \text{GM}^{2} = ac$, or $b = \text{HM}$ respectively, which order against the discriminant condition $b^{2} < ac$.