Problem: What is the largest $a$ for which all the solutions to the equation $3x^2+ax-(a^2-1)=0$ are positive and real?

Attempt: Solving the equation for $x$ I get $$x_{1,2}=-\frac{a}{6}\pm\sqrt{\frac{13a^2-12}{36}}.$$

For the roots to be real, the discriminant as to be greater than or equal to zero, so it yields the inequality

$$13a^2-12 \geq 0\Leftrightarrow -\frac{2\sqrt{39}}{13}\leq a\leq\frac{2\sqrt{39}}{13}.$$

Condition number two is that both roots should be positive. How should I think to proceed?

  • $\begingroup$ if the two roots are positive, $-b/2a$ is positive and $f(0)$ is also positive. $\endgroup$ – samjoe May 11 '17 at 8:39
  • $\begingroup$ How do you know this? $\endgroup$ – Parseval May 11 '17 at 8:42
  • $\begingroup$ @Paul Oh sorry clearly missed that! $\endgroup$ – samjoe May 11 '17 at 8:46
  • $\begingroup$ @samjoe My mistake actually. I thought the quadratic was positive definite. To OP; try finding the product of roots. $\endgroup$ – Twenty-six colours May 11 '17 at 8:49

First of all, the inequality you end up with is wrong: you need either $$a \leq -\frac{2\sqrt{39}}{13} \;\;\;\text{ or }\;\;\; a \geq \frac{2\sqrt{39}}{13}$$

Since the root $$x_2 = -\frac{a}{6} - \sqrt{\frac{13a^2 - 12}{36}} = \frac{1}{6}\left(-a - \sqrt{13a^2 - 12}\right)$$ is always smaller than $x_1$ (when both are real, of course), it suffices to find the largest $a$ such that $x_2$ is positive.

Hence we are looking for the largest $a$ such that $$-a - \sqrt{13a^2 - 12} \geq 0.$$

Now, as the root is always positive and minus the root is therefore negative, obviously we'll need $a \leq 0$. Rearrange as $$-a \geq \sqrt{13a^2 - 12}$$ and square both sides to find $$a^2 \geq 13a^2 - 12,$$ and rearrange again to find $$12 \geq 12a^2$$ or hence $$1\geq a^2.$$ Since $a$ must be negative, this means that $-1 \leq a \leq 0$.

Therefore, the values of $a$ such that the equation has $2$ (not necessarily distinct) real roots which are both positive, are $$a \in \left[-1,-\frac{2\sqrt{39}}{13}\right]$$ hence the largest such $a$ is $-\frac{2\sqrt{39}}{13}$

  • $\begingroup$ I like this solution! +1 $\endgroup$ – samjoe May 11 '17 at 8:48
  • $\begingroup$ 1) "Now, as the root is always positive and minus the root is therefore negative, obviously we'll need..." Why? 2) "Since $a$ must be negative, this means that..." Why? 3) How did you, with only a pen and a paper, find out that $$-1<\frac{2\sqrt{39}}{13} \ ?$$ $\endgroup$ – Parseval May 11 '17 at 9:41
  • $\begingroup$ 1) A square root is defined to be positive. So we have $$-a - (\text{ something positive} ) \geq 0,$$ or thus $$- a + (\text{ something negative }) \geq 0.$$ If $a$ were also positive, then $-a$ would also be negative, and the sum of two negative things is negative. So the entire left-hand side would be negative. $\endgroup$ – TastyRomeo May 11 '17 at 9:46
  • $\begingroup$ 2) if $1 \geq a^2$, then by taking the root of both sides, $1 \geq |a|$, or thus $-1 \leq a \leq 1$. But since $a$ is negative, we reduce this to $-1 \leq a \leq 0$. $\endgroup$ – TastyRomeo May 11 '17 at 9:47
  • $\begingroup$ 3) I simply followed your "style" of only having roots in the numerator. We were originally dealing with the inequality $13a^2 - 12 \geq 0$, or thus $a^2 \geq \frac{12}{13}$, and therefore $a \geq \sqrt{\frac{12}{13}} = \frac{2\sqrt{39}}{13}$ or $a \leq -\sqrt{\frac{12}{13}} = -\frac{2\sqrt{39}}{13}$. It's easy to see that $\sqrt{12/13} < 1$. $\endgroup$ – TastyRomeo May 11 '17 at 9:51

The conditions:

a) $\Delta \ge 0$ (the discriminant)

b) The sum of the roots is positive (Use Vieta)

c) The product of the roots is positive (Use Vieta)

  • $\begingroup$ What does the first triangle mean? $\endgroup$ – Parseval May 11 '17 at 8:51
  • $\begingroup$ @Parseval see the updated answer $\endgroup$ – user261263 May 11 '17 at 8:52
  • $\begingroup$ How did I miss this? :( $\endgroup$ – samjoe May 11 '17 at 8:56
  • $\begingroup$ So, the inequalitites I for b) and c) get should be b ) $$-a > 0 \Leftrightarrow a < 0$$ c) $$-(a^2-1) > 0 \Leftrightarrow a > 1 \ \text{or} \ a<-1$$ $\endgroup$ – Parseval May 11 '17 at 9:32
  • $\begingroup$ @Parseval Yes .... $\endgroup$ – user261263 May 11 '17 at 11:39

The equation is : $3x^2+ax-(a^2-1)=0$

  • First condition is which you identified, that the discriminant must be positive.

  • Note that the abscissa of vertex of this parabola is $\dfrac{-b}{2a}$ . For both roots to be positive, this value must be positive

  • Furthermore, since the parabola is upward, $f(0) > 0$ implies that both roots must be positive.

These are the three conditions which yield the sought answer.


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