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enter image description hereLet $p(x) = x^4+ax^3+bx^2+cx+d$, where $a,b,c, d$ are integers. The sum of the pair of roots are given by $1,2,5,6,9, 10$. find $p(\frac{1}{2})$.

I tried it by solving for roots and got $\alpha=-1.5, \beta=2.5, \gamma=3.5$ and $\delta=6.5$. But upon putting it in for $a,b,c$ and $d$,I got that $c$ and $d$ are not integers. What am I doing wrong?

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Feb 4 at 14:00
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    $\begingroup$ Please don't post the same question over and over. And please edit to show your work. We can't tell you where you went wrong if we don't see what you did. As a side note, there is more than one set of roots that give those pair sums. $\endgroup$
    – lulu
    Feb 4 at 14:00
  • $\begingroup$ Welcome to MSE, take a tour. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$
    – Marco
    Feb 4 at 14:30
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    $\begingroup$ @lulu Thanks for your response. I've added my word on this problem. As you've suggested that there exist more than one set of roots that give these sums then how should I approach this problem? Thank you. $\endgroup$
    – A4k45h
    Feb 4 at 15:52
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    $\begingroup$ @Math-fun Thanks for your response. I don't have the solution for this problem. Can you please support your answer for our better understanding? Thank you $\endgroup$
    – A4k45h
    Feb 4 at 15:55

2 Answers 2

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The comment from @lulu drived me on the right way.

Using your notations, $p(x)$ has 4 roots, let us call them $\alpha$, $\beta$, $\gamma$ and $\delta$, verifying $\alpha\le\beta\le\gamma\le\delta$. As all pair sums are distinct, we can even say that all roots are distinct so we have: $$\alpha\lt\beta\lt\gamma\lt\delta$$

It is evident that we have $\alpha+\beta = 1$ and $\gamma+\delta = 10$. But we cannot know a priori the order of $\alpha + \delta$ and $\beta+\gamma$

Knowing the sum of the lowest roots and the highest roots, we also know their average. So we can write:

$$\alpha = 0.5-h\\ \beta=0.5 + h\\ \gamma=5 - k\\ \delta=5+k$$ where $h\gt0$ and $k\gt0$

Again, $h+k = 3.5$ is evident, but we now have 2 possibilities, depending on the order of $\alpha + \delta$ and $\beta+\gamma$:

  • Hyp. 1: $\alpha + \delta\le\beta+\gamma$

    We get $h-k=0.5$, so $h=2$ and $k=1.5$ which gives your solution: $$\alpha=-1.5, \beta=2.5, \gamma = 3.5 \text{ and } \delta=6.5$$

  • Hyp. 2: $\alpha + \delta\ge\beta+\gamma$

    We get $k-h=0.5$, so $h=1.5$ and $k=2$ which now gives : $$\alpha=-1, \beta=2, \gamma = 3 \text{ and } \delta=7$$

It is now evident that $a$, $b$, $c$ and $d$ will all be integers...

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  • $\begingroup$ Excellent! I tried this problem for like $2.5$ hours but still didn’t got any clue on how to approach this. Thanks! $\endgroup$ Feb 4 at 19:03
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this is quite easily solved by creating a system of equations : $$p(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)$$

note that a, b, c and d are integers therefore $r_1, r_2, r_3, r_4$ too are integers. and since the sums are 1, 2, 5, 6, 9, 10 we have the sums of the roots being equal to 11 ( why? ).

so notice $$11 - r_1 - r_2 = r_3 + r_4$$ and also $$11 - r_3 - r_4 = r_1 + r_2$$ this is correct for other pairs as well. thus we arrive at the system below. $$ \{r_1 + r_2, r_3 + r_4\} = \{5, 6\} $$ $$ \{r_1 + r_4, r_3 + r_2\} = \{2, 9\} $$ and $$ \{r_1+r_3, r_2+r_4\} = \{1, 5\} $$ now notice that : $$ (r_1+r_2) - (r_1+r_3) + (r_3+r_2) = 2r_2 $$ so of the three terms above either all three must be even or two must be odd and one, even. therefore, we have either : $$ r_1+r_2 = 5 $$ and $$r_2 + r_3 = 2$$

or $$ r_1+r_2 = 6 $$ and $$r_2 + r_3 = 9$$ with $r_1 + r_3 \in \{1, 5\}$

to simplify we have : $$ r_1 - r_3 = \pm 3 $$ and $r_1 + r_3 \in \{1, 5\}$ thus $$(r_1, r_3, r_2, r_4) \in \{(2, -1, 3, 7), (4, 1, 1, 5), (-1, 2, 7, 3), (1, 4, 5, 1)\}$$

now to evaluate the polynomial : $p(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4) \Rightarrow p(\frac{1}{2}) \in \{\frac{63}{16}, \frac{-585}{16}\}$ But ! the $\{4, 1, 1, 5\}$ is unacceptable because there are no pairs that sum to 1, thus the answer shall be $\frac{-585}{16}$

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  • $\begingroup$ Good analysis, but I'm not sure how you got $(4,1,1,5)$ as a possible answer. No two of those sum to $1$, nor $10$. $\endgroup$
    – lulu
    Feb 4 at 16:52
  • $\begingroup$ oops, it is nearly 2 in the morning here i forgot to check for that! $\endgroup$
    – R TX
    Feb 4 at 16:58
  • $\begingroup$ No worries. The other solution works. $\endgroup$
    – lulu
    Feb 4 at 17:00

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