Find the values of m for which the equation $3x^2 − 2mx + 3 = 0$ has no solution.

Question

Is there a way to solve this algebraically without sketching the graph of the discriminant $(4m^2 - 36)$? What would the working out look like? Thanks in advance!

Answer 1

Recall the quadratic discriminant, $\Delta_2=b^2-4ac$. Note that if $\Delta_2 < 0$, there are no real roots to $f(x)=ax^2+bx+c$ since $\sqrt{\Delta_2}$ (which comes from the quadratic formula) must have $\Delta_2\geq 0$.

Thus, the values of $m$ for which $3x^2-2mx+3=0$ has no real roots are the solutions to $\Delta_2=4m^2-36 < 0 \implies m^2 < 9$. By the nature of parabolas, namely that $m = \pm 3$, the values of $m$ will be in the range $(-3, 3)$.

Answer 2

Here is a way, though I suspect it is better if you post your attempt so that someone may address the exact issue you have.

From AM-GM, $3x^2+3 \geqslant 6|x|$, and if $|m|< 3$, we have $2mx < 2|mx| < 6|x|$, so no solution is possible. OTOH, if $|m|\geqslant 3$, then when $x=\pm1$, in one of the cases the quadratic $3x^2+3 - 2mx$ turns negative, so the equation must have a solution.

Answer 3

The minimum of the function, taken in $x=\dfrac m3$, is equal to $-\dfrac{m^2}{3}+3$ and you want to have $-\dfrac{m^2}{3}+3\gt0$. Thus the possible values of $m$ should be such that $$-3\lt m\lt3$$