Matematické Fórum

stuart clark · 25. 01. 2012 20:08

Let $a$ and $b$ are the roots of the equation $x^2-10cx-11d=0$

and $c$ and $d$ are the roots of the equation $x^2-10ax-11b=0$ . Then find

(i) $a+b+c+d=$

(ii) value of $a\;\;,b\;\;,c\;\;d$

Where $a\;,b\;,c\;,d$ are Distinct Real numbers

Sulfan · 29. 01. 2012 14:32 — Editoval Sulfan (31. 01. 2012 19:23)

Autor skryl část textu. Zobrazit/skrýt!

vanok · 31. 01. 2012 15:44

Hi ↑ Sulfan:,

I have the other solutions
For example:
$(a;b;c;d)=(0;0;0;0)$ . In that case $a+b+c+d=0$ and two equations become $x^2=0$

I shall develop my solution when I shall more have time

Pavel Brožek · 31. 01. 2012 16:30

↑ vanok:

Hi, four zeros are not distinct real numbers :-). I think that Sulfan has found all solutions.

Sulfan · 31. 01. 2012 17:46

↑ Pavel Brožek:
Hi, you're right, in fact there are 4 possible solutions as I mentioned.

Except Vanok's $(a;b;c;d)=(0;0;0;0)$ also $(a;b;c;d)=(11;99;11;99)$ , but I understood under the term distinct numbers, that there shouldn't be any same pair of numbers.

Pavel Brožek

↑ Sulfan:

I think that the solutions with not distinct real numbers are $(a;b;c;d)=(0;0;0;0)$ and $(a;b;c;d)=(-11;-99;-11;-99)$ .

Edit: You didn't use Vieta's formulas correctly – the absolute term in quadratic equation is product of the roots without a minus sign.

Sulfan · 31. 01. 2012 19:22

↑ Pavel Brožek:
Thanks, it was my fault. I have edited my post.

stuart clark · 02. 02. 2012 05:09

Thanks to all,

but anyone explain me how can i get $a\;,b\;b\;,c\;d$

Thanks

Sulfan · 02. 02. 2012 14:22

↑ stuart clark: By solving the system of equation as an immediate result of Vieta's formula.

Autor skryl část textu. Zobrazit/skrýt!

vanok · 03. 02. 2012 12:33

Hi,
Finally, I have again the INTERNET connection
According to me, it is not enough to give answers without justification, but it is necessary to justify them.
Indeed, the solution that I quoted has ↑ Sulfan: do not answer at the request of ↑ stuart clark:, however to be able to answer his question, it seems to me to analyze, the problem in the possible more general case and then, to indicate solutions which he wishes to see.

On the other hand, I notice that the solution of ↑ Sulfan:, was revised with Viète's correct relations
However his solutions do not verify are not still right.

Then, I give here my analysis of this problem.
I am anxious to specify, that all the results are not obtained by equivalences so it is indispensable to verify that the solutions obtained verify the equations initially composed.(cf.↑ stuart clark:)

After this introduction, here is my attempt of solution of this problem (to criticize if need be)

Viète's relations of 2 equations are:
$a + b = 10c (V1) \\ ab = -11d (V2)\\c + d = 10a (V3)\\ cd = -11b (V4)$
Where from we obtain:
$abcd=121bd$ product $(V2)$ and $(V3)$
where still
$bd(ac-121)=0$
What gives us following three cases
$b=0$ or $d=0$ or $ac-121=0$

1° If $b=0$ Viète's relations become
$a = 10c \\ 0 = -11d \\c = 10a \\ cd = 0$
Finally
$a=b=c=d=0$

2° The symmetric case $d=0$ also looks
$a=b=c=d=0$

A remark, the cases $a=0$ and $c=0$ give too $a=b=c=d=0$

Remark:The solution $a=b=c=d=0$ give $a+b+c+d=0$ and two equation are
$x^2=0$

Let us suppose henceforth that $0 \notin \{a;b;c;d\}$

3° $ac-121=0$
give
$ac=121$ $(Rel1)$
$c=\frac {121}a$
Finally (provided that $a$ is effectively a root not zero of the initial equation)
$a+b+c+d=10(a+c)=10 $a+\frac {121}a $$
…………………………..
$(V1)$ give
$b= 10c-a$ $(V1’)$
and $(V3)$
$d=10a-c$ $(V3’)$
The first equation ↑ stuart clark: given has for root $a$ , so
$a^2 -10ac-11d=0$ , what give thanks to $(V3’)$
$a^2 -10ac-110a+11c=0$ $(EQ1’)$
Also, the second equation ↑ stuart clark:given has for root $c$ , so
$c^2-10ac-11b=0$ what give thanks to $(V1’)$
$c^2 -10ac-110c+11a=0$ $(EQ2’)$

$(EQ2’)$ - $(EQ1’)$ give,
$a^2-c^2 -121(a-c)=0$
Where from
$(a-c)(a+c-121)=0$
So we have to examine 2 cases
1° $a=c$
et
2° $a+c=111$ $(Rel2)$

Study of case 1°
$a=c$ , give with $ac=121$ $(Rel1)$ that $a \in \{-11 ;11\}$
For $a=c=11$ Viète's relations become
$11 + b = 110 \\ 11b = -11d \\11+ d = 110 \\ 11d = -11b$
What drives to a contradiction.

For $a=c=-11$ Viète's relations become

$-11 + b = -110 \\ -11b = -11d \\-11+ d = -110 \\ -11d = -11b$
What gives:
$a=c=-11\\b=d=-99$
And in that case there both equations of departure become
$x^2+110x+1089=0$
and
$a+b+c+d=-210$

Study of case 2°
$a+c=121$ $(Rel2)$
with
$ac=121$ $(Rel1)$
show that has $a;c$ are solutions of the equation
$x^2-121x+121=0$
Thus
$\{a;c\}=\{ \frac{11*(11-3\sqrt{13})}2;\frac{11*(11+3\sqrt{13})}2\}$
Viète's relations :
$a + b = 10c (V1) \\ c + d = 10a (V3))$
allow to express $b;d$ corresponding

But we obtain a contradiction with $( V2); ( V4)$

CONCLUSION:
The problem admits only two solutions in \mathbb{R^4},
1) $(a;b;c;d)=(0;0;0;0)$
with $a+b+c+d=0$ and two equation are
$x^2=0$
and
2) $(a;b;c;d)=(-11;-99;-11;-99)$
and in that case there both equations of departure become
$x^2+110x+1089=0$
with
$a+b+c+d=-210$

and anybody which satisfied by the announced requirements.