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Question

Find the value of $a$ such that equation $$f(x)=x^2+(a-3)x+a=0$$ has exactly one root $\alpha$ between the interval $(1,2)$ and $f(x+\alpha)=0$ has exactly one root between the interval $(0,1)$.

Attempt

Discriminant$=0$ for exactly one root.

$F'(x)=0$ where $x$ will lie between $(1,2)$ and hence another restriction on a and $\alpha$. But how will I implement it on second part of the question $f(x+\alpha)$??

Any hints and suggestions are welcome.Its question number 5. enter image description here

jayant98
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  • For clarification : Is the $a$ on the expression different than the $\alpha$ of the root ? – Rebellos Nov 09 '18 at 11:11
  • Yes. You got it right. – jayant98 Nov 09 '18 at 11:12
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    Note that "Exactly one root in the interval $(1, 2)$" does not mean that it has only one root (and therefore does not mean that the discrtiminant is $0$). It means what it says: That between $1$ and $2$, there is exactly one root, but there may be another root elsewhere. – Arthur Nov 09 '18 at 11:13
  • @Arthur So then it will have two cases to present that 2 roots case where one root is in given interval and the 1 root case. Am I right? – jayant98 Nov 09 '18 at 11:14
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    @jayant98 Yes. You will, however, find that "$f(x + \alpha) = 0$ has exactly one root in the interval $(0,1)$" implies that there are indeed two distinct roots. – Arthur Nov 09 '18 at 11:15
  • Now discriminant will be greater or equal to zero. But the main thing I want to know is how will I immplement the 2nd part of the question? – jayant98 Nov 09 '18 at 11:17
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    I have now looked at the problem, and this is the point where I ask you whether you are certain that you've copied everything correctly. Are you certain that the intervals are $(1, 2)$ and $(0, 1)$? Are you certain that it's $f(x + \alpha)$? Because I can't find any such $f$. See this geogebra graphing to see what I mean. Here you can adjust the value of $a$, and at no point willl $g(x) = f(x + \alpha)$ have a positive root. – Arthur Nov 09 '18 at 11:32
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    @Arthur If you look to the other solution for alpha and plug $a=0.95$ it works – Bruno Andrades Nov 09 '18 at 13:45
  • @BrunoAndrades But then $\alpha\notin (1, 2)$. So that doesn't work. – Arthur Nov 09 '18 at 13:48
  • @Arthur $\alpha$=1.3422.. is actually in (1,2). But $f(x+\alpha)=0$ has no solution in (0,1) – Damien Nov 09 '18 at 14:19
  • @Arthur,@Bruno,@Damien I can give you the picture of it. – jayant98 Nov 09 '18 at 17:38
  • Wait, I should just upload the image. – jayant98 Nov 09 '18 at 17:40

2 Answers2

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Write $$a = \underbrace{3x-x^2\over +x+1}_{=g(x)} $$

Now we are searching for $a$ if $y=a$ cuts exactly once $g(x)$ at $\alpha$ which is in interval $(1,2)$ and graph $g(x+\alpha)$ in interval $(0,1)$. If we mark $y=x+\alpha$ we see that $-1<y<0$. If we draw a graph of $g$ we see that $y=a$ can not cut graph of $g$ at the same time in $(-1,0)$ and $(1,2)$:enter image description here

nonuser
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There is no solution.

f is a second order polynomial so having only one solution in (1,2) means that either

(1) $f(x) = (x-\alpha)^2$ or

(2) $f(2)$ and $f(1)$ have different signs.

(1) implies $a =\alpha = 1$ which is not in (1,2)

(2) $f(2) = 3a-2$ and $f(1) = 2a-2$ so $f(2)f(1) = (3a-2)(2a-2) = 6(a-2/3)(a-1)$ and therefore, $f(2)f(1)<0 <=> 2/3<a<1$

Thus $a$ has to be in $(\frac{2}{3},1)$.

But $αβ=a<1$ and $α>1$ so $β<1<α$ with $β$ being the second root of $f$. So $f(x)>0$ for $x>α$ and $f(x+α)>0$ for $x>0$.

To sum up,

$f$ has one root in $(1,2)$ $\implies$ $a\in(\frac{2}{3},1)$ $\implies$ $x \mapsto f(x+\alpha)$ has no root in $(0,1)$

Damien
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    If $a = \alpha = 1$ and $f(x) = (x-1)^2$, then $f(x+\alpha)$ has only one root: $x\in 0$. That's not contained in $(0,1)$. So it doesn't verify the sevcond restriction. And it doesn't verify the first either, as $1\notin (1, 2)$. Same goes for your final line: $0\notin (0,1)$. – Arthur Nov 09 '18 at 13:50
  • @Arthur: I believe that 0 is actually in [0,1] and 1 is in [1,2] as well. What does (1,2) means to you? ]1,2[ ? – Damien Nov 09 '18 at 13:57
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    Yes, $(1, 2)$ means $]1, 2[$, and is actually a lot more common; I have only ever seen the $]1, 2[$ notation in a couple of posts here and there on this forum, and nowhere else. – Arthur Nov 09 '18 at 13:58
  • OK, I didn't know about it. It might not be very popular notation in France. – Damien Nov 09 '18 at 14:03
  • @Damien How did you implemented that 2nd part $f(x+\alpha)=0$? I mean for every value of a there will be different $\alpha$ and how did you put or used that alpha in that equation to put forward the root's location? – jayant98 Nov 09 '18 at 18:17
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    @jayant98 $\alpha\beta=a<1$ and $\alpha>1$ so $\beta<1<\alpha$ with $\beta$ being the second root of f. so f(x)>0 for $x>\alpha$ and $f(x+\alpha)>0$ for x>0 – Damien Nov 10 '18 at 13:33
  • @Damien Why don't you edit this last comment in your answer? I shall then upvote it for the answer. Thanks for the help! – jayant98 Nov 13 '18 at 20:11