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If a root of the equation $$3x^2 + 4x + 12a + 9ax = 0$$ is greater than 6, then the correct statement of the coefficient $a$ is:

a) $a = 2$

b) $a> -2$

c) $a = -2$

d) $a <-2$

e) $-2\le a\le 2$

What I did was solve it as if a root was exactly $6$, and I find that $a$ is equal to $2$ ... But the premise says that a root has to be greater than $6$, not equal to $6$ ... and I do not know how find that, for now I would know that $a$ does not have to be $2$ ...

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  • $\begingroup$ What is the reason for the downvote? $\endgroup$ – mfl Aug 10 '18 at 18:23
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    $\begingroup$ @mfl I suspect it has something to do with the question being written in Spanish. $\endgroup$ – Mastrem Aug 10 '18 at 18:25
  • $\begingroup$ Why did you choose to write this post in Spanish? Reading your other posts it seems that you are perfectly capable of writing in English. $\endgroup$ – TheSimpliFire Aug 10 '18 at 18:34
  • $\begingroup$ If $a = 2$, then the roots must be negative. Did you mean $a = -2$? $\endgroup$ – N. F. Taussig Aug 10 '18 at 18:45
  • $\begingroup$ Solve $3x^2+4x+12a+9ax = 0$ for $a$ to get $a = -\frac{x}{3}$ (note the solve step assumes $x\not= -\frac{4}{3}$ when we divide by $9x+12$, but you are next going to assume $x>6$ so this case is not a problem) $\endgroup$ – Winther Aug 10 '18 at 20:44
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Hint: $\;3x^2 + 4x + 12a + 9ax = x(3x+4)+3a(3x+4)= (x+3a)(3x+4)\,$.

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Hint

\begin{align} 3x^2+4x+12a+9ax = 0&\iff 3x^2+(4+9a)x+12a= 0\\ &\iff x=\dfrac{-4-9a\pm \sqrt{(4+9a)^2-144a}}{6}.\end{align} So,

$$\dfrac{-4-9a\pm \sqrt{(4+9a)^2-144a}}{6}\ge 6 \iff -4-9a\pm \sqrt{(4+9a)^2-144a}\ge 36. $$

That is

$$-4-9a\pm \sqrt{16+72a -63a^2}\ge 36$$

So, you can check what is the correct answer.

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  • $\begingroup$ Llego a que $\;2a^2 + 9a + 22<0,$ $\endgroup$ – Strong Aug 10 '18 at 22:57
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Using @dxiv hint $(x+3a)(3x+4) = 0$ we have $x = -\frac{4}{3} < 6$ and $x = -3a > 6$ Therefore, $a < -2$

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