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Q: Find the values of m for which the expression below is always positive. $x^2 + 2mx + (3m-2)$

I have attempted the question and know that I'm supposed to use the discriminant, however I'm having a bit of trouble with the substitution. I factorised first to get $4(m^2x^2-3m+2)$ then tried factorisation by grouping but I couldnt get it. Any ideas?

$$\begin{align}x^2+2mx+(3m-2)\\b^2-4ac>0\\(2mx)^2-4*1*(3m-2)\\4m^2x^2-12m+8=0\\4(m^2x^2-3m+2)=0\end{align}$$ ??

Thanks for your help! :)

Zach
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    Have you heard of this thing called a discriminant? It might be useful. Show your work, this is not a 'Solve-My-Homework' site – Mathematician 42 Feb 15 '18 at 09:39
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    Hey buddy. thanks for the kind words. Rather than be an internet troll how about you provide an answer for the purpose of the site. No need to insult people for their mathematical ability, I'm purely asking a question which I'm not sure how to do. Any help? :) – Zach Feb 15 '18 at 09:44
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    Hey @Zach , while maybe a bit harshly worded, I think Mathematician 42 has a bit of a point. Usually here you'll get better responses if you show some of your work (or if you really have no idea how to get started then tell us that). Most people will try to guide you through the question rather than just give an answer, after all we are all here to learn! – TheMathsGeek Feb 15 '18 at 09:49
  • Zach, notice that @Mathematician42 actually provided you with a very useful hint. Can you please read his/her comment more carefully instead of resorting to insults yourself? –  Feb 15 '18 at 09:49
  • I've read the comment, and attempted the question but I didn't get it. The reason I tagged the discriminant is I know that's what you are supposed to do, however I wasn't successful. Please allow me to clarify :) – Zach Feb 15 '18 at 09:51
  • Edited my post, any ideas – Zach Feb 15 '18 at 09:53
  • You could post (1) have you calculated the discriminant and what formula you've obtained (maybe you just made a mistake in your calculation), and also (2) whether you know how the value of the discriminant affects whether a quadratic function is always positive or not. Then, we can see how to connect the dots. –  Feb 15 '18 at 09:53
  • Okay, I will attach all working out – Zach Feb 15 '18 at 09:54
  • Sure, thanks for the edit @Zach. Do you know that the discriminant tells you about the roots of the polynomial? So if the discriminant D < 0 the function will never have real roots (so is either always positive or always negative - it never touches the axis). Then you need to transfer this to a statement about m. If you're still confused let me know! – TheMathsGeek Feb 15 '18 at 09:56
  • Ah, now I get it where you are making the mistake... $x^2+2mx+3m-2$ - you need to view it as $ax^2+bx+c$. What are $a,b,c$? Let me write it a bit differently: $x^2+2mx+3m-2=1\cdot x^2 + (2m)\cdot x + (3m-2)$... Note 1$\cdot x^2$. –  Feb 15 '18 at 09:56
  • Isn't a=x^2, b=2mx and c=(3m-2)? – Zach Feb 15 '18 at 09:59
  • Can someone please post working out so I can learn how to do it? – Zach Feb 15 '18 at 10:00
  • @Zach Nope, see my answer. $a,b,c$ have no '$x$' in them, they are the numbers that multiply $x^2$, $x$ and $1$, respectively. –  Feb 15 '18 at 10:02
  • okay I see, let me try – Zach Feb 15 '18 at 10:02
  • That escalated quickly (insert meme here). My words weren't meant as personal attack. Take it as useful hint for future questions. Also, try to learn the basics of Mathjax (a handy code for formatting mathematical questions). Your questions will look a thousand times better and more people are likely to help you. – Mathematician 42 Feb 15 '18 at 10:02
  • Nws, cant see facial expressions and intent through words :) – Zach Feb 15 '18 at 10:04
  • I got (2m)^2-41(3m-2) then 4m^2-12m+8, then 4(m^2-3m+2), then 4(m-2)(m-1) Is that right? – Zach Feb 15 '18 at 10:05
  • Well done. Now try the rest, and let us know, but please with plenty of details, just like a minute ago, if you get stuck. –  Feb 15 '18 at 10:06
  • Is my answer above correct? – Zach Feb 15 '18 at 10:08
  • The discriminant is correct, but your problem was not to calculate the discriminant. Your problem was to use the discriminant to determine when the function $x^2+2mx+3m-2$ will be always positive, if I remember correctly. –  Feb 15 '18 at 10:09
  • Never mind I didn't get it. I tried to figure it out but its too hard. Thanks for your help guys. :( – Zach Feb 15 '18 at 10:12
  • Come on... Don't give up! When is $ax^2+bx+c$ always positive for every $x$? - You must have learned that at some point; look through your book/notes... –  Feb 15 '18 at 10:13
  • ax^2+bx+c is positive when the discriminant has 2 solutions, 1 solution for x=1 and no solutions for x<0 – Zach Feb 15 '18 at 10:15
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    Well, it is a bit different. When the discriminant is positive, $ax^2+bx+c=0$ has two solutions, so it changes sign from positive to negative and back to positive (or vice versa). If the discriminant is zero, then $ax^2+bx+c$ has one solution, and it stays positive (or negative) all the time, except at that point, where it is zero. If the discriminant is negative, then $ax^2+bx+c=0$ has no solutions, and $ax^2+bx+c$ stays always positive or always negative. (Always positive if $a\gt 0$). –  Feb 15 '18 at 10:27
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    (Cont'd) This is easiest seen if you've gone through examples and drawn the curves for $ax^2+bx+c$ in the coordinate plane, for different $a,b,c$ - not sure if you've done that in class. Basically, in your problem you don't want $ax^2+bx+c$ to have any zeros, it has to stay always positive! What does that mean for the discriminant? –  Feb 15 '18 at 10:29

3 Answers3

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Hint: If you want to represent $x^2 + 2mx + (3m-2)$ as $ax^2+bx+c$, then:

$$a=1, b=2m, c=3m-2$$

so the discriminant is:

$$b^2-4ac=(2m)^2-4\cdot 1\cdot (3m-2)$$

Hope you can take it from here.

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$$x^2 + 2mx + (3m-2)=(x+m)^2-(m^2-3m+2)$$

Since $(x+m)^2 \ge 0$, all you need is to solve $$m^2-3m+2 < 0$$

Jaideep Khare
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  • (m-2)(m-1)<0, so m= 1 or 2? – Zach Feb 15 '18 at 10:16
  • Also, it matters a lot whether the original problem statement is $x^2+2mx+(3m-2)\gt 0$ or $\ge 0$ for every $x$. (It says "$x^2+2mx+(3m-2)$ is always positive": does $0$ count as positive, that is...) @Zach, can you please clarify? –  Feb 15 '18 at 10:41
  • @Zach $(m-1)(m-2)<0$ so, $m \in (1,2)$, i.e. all values between $1$ and $2$, excluding $1$ and $2$. – Jaideep Khare Feb 15 '18 at 10:41
  • @user8734617 Positive mean $>0$ , so I think it's clear, now. (see the first line of question) – Jaideep Khare Feb 15 '18 at 10:42
  • It means the same to me. (See my earlier comments to the question itself.) However, you know, some teachers out there may teach differently, and I wanted to be 100% sure we don't give @Zach a bad advice... –  Feb 15 '18 at 10:43
  • @user8734617 I don't understand what you saise "It means same to me"... Do you mean positive $\implies \ge 0$ – Jaideep Khare Feb 15 '18 at 10:45
  • No, no, I mean, I also, when seeing the word "positive", assume it means $\gt 0$. I was just not sure what Zach's teacher (or whoever set up the problem) assumes. –  Feb 15 '18 at 11:09
  • @user8734617 Oh I understand now. You mean to say it may depend on the person who formatted this question to assume $0$ in positive numbers. – Jaideep Khare Feb 15 '18 at 19:26
  • @JaideepKhare Exactly. Just like $\mathbb N$ for some people contains $0$ and for some people doesn't. (However, sadly, we haven't had any feedback from the OP, so in all likelihood they have given up with this question...) –  Feb 15 '18 at 19:28
  • Hey, positive doesn't include zero – Zach Feb 17 '18 at 06:53
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Since $$4(m^2x^2 - 3m + 2)= 0$$ then because $0=0\times 4$, we can actually write that $$m^2x^2 - 3m + 2 = 0.$$ And since this equation is a quadratic equation (particularly a quadratic trinomial) with only one variable we want to solve for, namely $m$ (here, $m$ is known is the indeterminate) then we can use the quadratic formula to solve for $m$.

Quadratic Fromula: Provided the equation $ax^2 + bx + c = 0$ for constants $a, b, c, x$ with $a\neq 0$, $$x=\frac{-b\pm\sqrt{\Delta}}{2}$$ such that $\Delta = b^2 - 4ac =$ the discriminant. Proofs of this formula can be found here.

Now simply substitute.

Mr Pie
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  • Got it thanks :) – Zach Feb 15 '18 at 10:22
  • @Zach No problem :)) – Mr Pie Feb 15 '18 at 10:26
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    @Zach's determinant is not $m^2x^2-3m+2$ - it cannot depend on $x$ because the original quadratic formula is $x^2+2mx+(3m-2)$ so there isn't any point in looking into this expression at all! This is a distraction at best. –  Feb 15 '18 at 10:32