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If $a, b$ are the roots of the equation $x^2-10cx-11d=0$ and $c,d$ are the roots of the equation $x^2-10ax-11b=0$ (where $a\ne b\ne c\ne d\ne 0$), then find the value of $a+b+c+d$.

I have the following data:

  • $a+b=10c$

  • $c+d=10a$

  • $ab=-11d$

  • $cd=-11b$

  • $\frac{c-a}{b-d}=\frac1 {11}$

Tejas
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    Facts 3&4 let you conclude that $ac=121$. – vadim123 Nov 17 '13 at 04:27
  • @vadim123 with some additional please-dont-be-zero work – chubakueno Nov 17 '13 at 05:14
  • What is interesting is that they gave you the fifth data. In principle, there are four unknwowns then four equations (data) should be sufficient. In your opinion, what is the role of the fifth data ? – Claude Leibovici Nov 17 '13 at 06:01
  • As the equations aren't linear, it's possible that four equations are not enough. – Doc Nov 17 '13 at 06:18
  • @Doc. Four equations for four unknowns have solutions (multiple, may be). – Claude Leibovici Nov 17 '13 at 06:46
  • @ClaudeLeibovici, that is the data which I have evaluated. They have given only the equations. – Tejas Nov 17 '13 at 14:59
  • @Claude, of course you are correct. I meant that four equations may not be enough to determine a unique solution. It's clear that my phrasing was very poor. – Doc Nov 18 '13 at 15:16
  • @Doc. This is why they gave the fifth relation; it ensures unicity. – Claude Leibovici Nov 19 '13 at 16:38
  • @Claude, but it doesn't. The fifth relation is derived directy from the first two relations. Thus there are multiple solutions to the system of five relations if and only if there are multiple relations to the first four. – Doc Nov 19 '13 at 20:29

1 Answers1

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Only $c\neq a$ is needed, as shown below.

Let $p=10$ and $q=11$. Then we have

$$ x^2-pcx-qd=(x-a)(x-b) \\ x^2-pax-qb=(x-c)(x-d) \tag{1} $$

so that

$$ a+b=pc, ab=-qd, c+d=pa, cd=-q \tag{2} $$

We deduce that

$$ b=pc-a, d=pa-c \tag{3} $$

so that (2) gives $a(pc-a)=-q(pa-c)$ and $c(pa-c)=-q(pc-a)$, and hence

$$ a^2=(pq)a-(q)c+(p)ac,\\ c^2=(pq)c-(q)a+(p)ac \tag{4} $$

Substracting, we see that $c^2-a^2=(pq)(c-a)+q(c-a)$ and hence $a+c=pq+q=q(p+1)$. Finally we have

$$ a+b+c+d=a+c+(pc-a)+(pa-c)=p(a+c)=p(p+1)q $$

In your example, one obtains $a+b+c+d=10\times 11 \times 11=1210$.

Ewan Delanoy
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