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1.find a for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ such that for every m real, f has real roots 2.find m such that for every a real, f has real roots

My ideea is to demonstrate that $\delta=4(m-a)^2 - 4(3am-2) = 4(m^2-5am+a^2+2) >0$ But I have no ideea how to do this.

oren revenge
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  • For 1) treat $m$ as a constant and solve your inequality for $a$ in terms of $m$, and for 2) do the reverse. – Avi May 27 '16 at 19:20

2 Answers2

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As you correctly remark, it is enough that

$$\delta=4(m^2-2am+a^2)-4(3am-2)\ge0\iff a^2-5ma+m^2+2\ge0$$

Now look at the discriminant of the new quadratic in $\;a\;$ you got:

$$\delta_a:=25m^2-4m^2-8=21m^2-8$$

We'd like to check when $\;\delta_a\le0\;$ so that that $\;\delta\ge0\;$. This will be true for all $a$ iff

$$m^2<\frac8{21}\iff |m|\le\frac{2\sqrt2}{\sqrt{21}}$$

Taking into account that the quadratic $\;\delta\;$ is symmetric with respect to both $\;a,m\;$ we'll get the same result if we fix $\;a\;$ and calculate $\;\delta_m\;$ instead: $f(x)$ will have real roots for all $m$ iff $$|a|\le\frac{2\sqrt2}{\sqrt{21}}$$

almagest
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DonAntonio
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  • I don't understand your final comment. If $|a|$ is sufficiently small then $f(x)$ will have real roots for any value of $m$ (as we can easily check for $a=0$). Similarly, if $|m|$ is sufficiently small, then $f(x)$ will have real roots for any value of $a$ (as again we can easily check for $m=0$). – almagest May 27 '16 at 19:41
  • @almagest Thank you. From what I understood in the question, it is required to find the values of $;a;$ such that the original quadratic has real roots for all $;m;$ . Of course one can, as shown, get real roots for $;|m|;$ small enough, but not for all the possible values. The same thing happens with the roles of $;a,m;$ changed because of the symmetry shown in the original quadratic's discriminant. – DonAntonio May 27 '16 at 19:48
  • It says 1. Find $a$ such that it has real roots for every $m$, 2. Find $m$ such that it has real roots for every $a$. – almagest May 27 '16 at 19:49
  • @almagest Exactly. In (1), for example, for what values of $;a;$ will $;\delta\ge0;$ ? If and only if $;m;$ is such that $;|m|<\frac{2\sqrt2}{\sqrt{21}};$ , which means: there is no value of $;a;$ such that for all values of $;m;$ the quadratic has real roots. Perhaps I confused something here? – DonAntonio May 27 '16 at 19:51
  • If $a=0$ then it has real roots for ALL values of $m$. Similarly for all sufficiently small $a$. – almagest May 27 '16 at 19:53
  • @almagest Right. I messed this up big time. Thank you, deleting and, perhaps, later editing. – DonAntonio May 27 '16 at 19:57
  • NO. It is almost right. You just need to slightly change it at the end! – almagest May 27 '16 at 19:58
  • Drats! They tell me I can't delete an accepted answer...ok, I hope people sees this comment interchange, and later I'll try to fix things. – DonAntonio May 27 '16 at 19:58
  • try now to modify – oren revenge May 27 '16 at 19:58
  • @almagest Oh, I think I see what you mean....:) . But I need to run now. Shall be back later. Thank you very much – DonAntonio May 27 '16 at 19:59
  • @Joanpemo I have made a couple of small changes to your answer which I think fix things. Please check. :) – almagest May 27 '16 at 20:26
  • @almagest Ok, I'm back. Thank you very much for your edition. Exactly what I thought after reading your comments and after I messed up something that simple. Very nice. – DonAntonio May 27 '16 at 21:20
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A parabola $Ax^2+Bx+C$ with $A>0$ attains its minimum at $x_0 = -\frac{B}{2A}$, with value $y_0 = C-\frac{B^2}{4A}$. First the first question, this means that the minimum $\mu_a$ of $(m^2−5am+a^2+2)$ is always positive, whatever $m$. You can expect some solutions: if $a=0$, this quantity is equal to $\mu_0=m^2+2$, positive for every $m$.

The minimum value in general is $\mu_a =a^2+2 -\frac{(5a)^2}{4}=\frac{8-21a^2}{4}$. If $|a|\le \sqrt{\frac{8}{21}}$, then the quantity $\delta(m,a)$ is positive.

A verification: if $a=0$, the discriminant of $f_0(x) = x^2+2mx−2$ is $\delta_0 = 4m^2+8$ which is always positive, for all $m$.

You can use the same approach the other way round for the second question. But you really don't need: $\delta(m,a) = \delta(a,m)$, the function is symmetric with respect to the two variables. So you should find a quite similar result for $m$.

Laurent Duval
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