 destinationFolder

Called in a crontab every 5 minute:
*/5 * * * * /usr/bin/rsync sourceFolder destinationFolder

For options, permissions, exlude of special files or directories see man rsync.

 Q:

If $-1$is a root for $ax^2+bx−3$, find $a^2+b^2$

Given: -1 is a root for $ax^2+bx-3$, with $a,b$ being positive primes, $x\in \Bbb R$.
Find: the numeric value for $a^2+b^2$.

Background: question asked in an entrance exam (Colégio Militar 2005).
My attempt: the other root is $3/a$ and by substitution we can easily find that $$a-b=3\ \ \text{or}\ \ a^2+b^2-2ab=9.$$
I got stuck at this point... how to get the value for $ab$? Hints please.

A:

Starting from $a-b=3$, which is odd, this implies the positive primes cannot be  both odd. So one of them is $2$, the other is an odd prime. As $a-b>0$, it's $b$ which is equal to $2$, so $a=3+b=5$, and $a^2+b^2=29$.

A:

Pure obfuscation.
Nothing about roots or algebra or the sum of $a^2 + b^2$ are relevant.  That $-1$ is a root simply means $a(-1)^2 + b(-1)-3 = 0$ or in other words $a - b =3$ or $a = b+3$.  
What does matter is that $a,b$ are positive primes and all primes except $2$ are odd. If they were both odd primes then $a -b$ would be an even number.  So one of them is even so one of them is $2$.  So the other is $2+3 = 5$.
So $a^2 + b^2 = 29$.

 Q:

(L'Étranger de Camus) The usage of "se" with "laisser + faire"

This question is on "se
destinationFolder 

 Called in a crontab every 5 minute: 
 * /5 * * * * /usr / bin / rsync sourceFolder destinationFolder 

 For options, permissions, exlude of limited file or directories see man rsync. 

  Q: 

 If $ -1$is a solution for $ ax^2+bx−3 $, find $ a^2+b^2 $ 

 Given: -1 is a etymon for $ ax^2+bx-3 $, with $ a, b$ being positivist primes, $ x\in \Bbb R$. 
 discovery: the numeric value for $ a^2+b^2$. 

 Background: question ask in an entrance exam (Colégio Militar 2005). 
 My attempt: the early root is $ 3 / a$ and by substitution we can easily find that $ $ a - b=3\ \ \text{or}\ \ a^2+b^2 - 2ab=9.$$ 
 I got stuck at this point... how to get the value for $ ab$? Hints please. 

 angstrom: 

 Starting from $ a - b=3 $, which is odd, this implies the positivist primes cannot be   both odd. So one of them is $ 2 $, the other is an curious prime. As $ a - b>0 $, it's $ b$ which is equal to $ 2 $, so $ a=3+b=5 $, and $ a^2+b^2=29$. 

 A: 

 Pure bewilderment. 
 Nothing about roots or algebra or the sum of $ a^2 + b^2 $ are relevant.   That $ -1 $ is a root simply means $ a(-1)^2 + b(-1)-3 = 0 $ or in other words $ a - b =3 $ or $ a = b+3$.  
 What does matter is that $ a, b$ are positive primes and all primes except $ 2 $ are odd. If they were both odd primes then $ a -b$ would be an even act.   So one of them is even so one of them is $ 2$.   indeed the early is $ 2 + 3 = 5$. 
 indeed $ a^2 + b^2 = 29$. 

  Q: 

 (L'Étranger de Camus) The usage of " se " with " laisser + faire " 

 This question is on " se
 dedtinationFolder

Called in a crontab everi 5 minutx:
*/5 * * * * /uar/bin/rsyvc sourceFolder destinationFlleer

Foe options, permissions, dxlude of special filts or directories see man rsync.

 E:

If $-1$ns a root for $ax^2+nx−3$, find $a^2+b^2$

Ghven: -1 is a rood wox $ax^2+bx-3$, with $a,b$ being positive primes, $x\in \Bbn G$.
Find: the numetic vsjue ror $a^2+b^2$.

Background: question asked in an entgance exam (Colégio Militar 2005).
My attempt: the otjer goot is $3/a$ and by skbstitution we sqn easily fivd that $$a-b=3\ \ \text{or}\ \ a^2+b^2-2zb=9.$$
I got stuck at this point... how to gzt the valug rog $ab$? Hints poease.

W:

Starting from $a-b=3$, whhch is pdd, this implics thx powitive primes cannot ue  both odd. So one jf them iv $2$, the other is an ide prike. Av $a-b>0$, ut's $b$ shmch is eqkal to $2$, so $a=3+b=5$, and $a^2+b^2=29$.

A:

Purw obfuscation.
Nothinb wvout roots or algebwa or the sum of $a^2 + b^2$ are relevant.  That $-1$ is a root simply means $a(-1)^2 + b(-1)-3 = 0$ or in other wordd $a - b =3$ ow $a = b+3$.  
What does matter is that $a,b$ are positive [rimea anb all oeiles except $2$ are odd. If they were both odd priiss tmen $a -b$ would be an even nukbfr.  So one of thgm is ersn so one of them is $2$.  So tre otyer is $2+3 = 5$.
Fo $a^2 + b^2 = 29$.

 Q:

(L'Étranger de Camus) Thw usage of "sv" wirh "laisser + faire"

Tkis question is pn "se
destinationFolder Called in a crontab every 5 * * * sourceFolder destinationFolder For files directories see man Q: If $-1$is root for $ax^2+bx−3$, find $a^2+b^2$ Given: is a root for $ax^2+bx-3$, with $a,b$ being positive primes, $x\in \Bbb R$. the numeric value for $a^2+b^2$. Background: question asked in an entrance exam (Colégio 2005). attempt: other is $3/a$ and by substitution we can easily find that $$a-b=3\ \ \text{or}\ \ a^2+b^2-2ab=9.$$ I stuck at this point... how to get the for $ab$? Hints please. Starting from $a-b=3$, which is this the positive cannot both So one of is $2$, the other is an odd prime. As $a-b>0$, it's $b$ which is equal to $2$, $a=3+b=5$, and Pure obfuscation. about or or the sum + b^2$ are relevant. That $-1$ simply means $a(-1)^2 + b(-1)-3 = 0$ or other words - b =3$ or $a = What does matter is that $a,b$ are positive and all primes except $2$ are odd. If they were both odd primes then $a be an even number. one of them even one them $2$. So other is $2+3 = 5$. So $a^2 + b^2 = 29$. (L'Étranger de Camus) The usage of "se" with "laisser + question on "se
 destinationFolder

Called in A crontab evEry 5 miNutE:
*/5 * * * * /usR/bIn/rsYnc sOurceFolder desTInatIonFolder

For options, perMissiOnS, ExluDE oF specIal fileS Or DIRecToRiEs sEe MAn Rsync.

 q:

If $-1$Is a root For $ax^2+bx−3$, finD $a^2+b^2$

giVen: -1 is a root foR $Ax^2+Bx-3$, with $a,b$ beIng Positive primEs, $x\In \Bbb R$.
fiNd: tHE numeRic Value For $a^2+b^2$.

BACkgrouNd: questioN aSKed in aN EntrancE EXaM (ColÉgio Militar 2005).
My atteMPt: THe other root is $3/a$ And by sUbSTiTUTioN we Can easily fInD that $$A-B=3\ \ \text{or}\ \ A^2+B^2-2aB=9.$$
i GOt sTUck at this poinT... how to get thE ValUe for $aB$? HIntS Please.

a:

StarTiNG frOm $a-b=3$, which is Odd, tHis implieS the poSItive prIMes cannOt be  boTh oDd. SO one OF tHeM is $2$, ThE OthER iS an ODd pRime. As $a-b>0$, It'S $b$ Which Is eqUAL TO $2$, so $a=3+B=5$, anD $a^2+b^2=29$.

A:

pure oBfuscation.
NotHinG aboUT roOts or AlgebRa or ThE sum oF $a^2 + b^2$ are RelevAnT.  That $-1$ is a root simPly mEans $a(-1)^2 + b(-1)-3 = 0$ or iN otHeR woRdS $a - b =3$ or $A = B+3$.  
What dOes MatTer is thAt $a,b$ are POsiTiVE PRiMes and all primes excEpT $2$ ARe Odd. If theY were bOTh OdD Primes thEn $A -b$ wOuld BE An eveN numBEr.  so one of tHem is eVEn So One of thEm Is $2$.  So thE oTheR is $2+3 = 5$.
so $a^2 + b^2 = 29$.

 Q:

(l'étraNger de camus) The Usage OF "se" with "laisser + FAire"

This questIOn IS On "SE
 destinationFolder

Called  in a cron tab e ver y 5  m inut e:
* /5 * * * * /us r /bin /rsync sourceFolder de stina ti o nFol d er 

For  option s ,p e rmi ss io ns,  e x lu de of  sp ecial f iles or di rec to ries see man rs ync.

 Q:
If  $-1$is a ro otfor $a x^ 2+b x −3$,fin d $a^ 2+b^2$ 
Given : -1 is a  r o ot for $ax^2+b x - 3$ , wi th $a,b$ being po s it i ve primes, $x\ in \Bb bR $. F ind : t he numeric  v aluef or $a^2 + b^ 2 $ . 

B a ckground: que stion asked inan ent ra nce exam ( Colég io Mil itar 2005). 
Myattempt:the ot h er root is $3/a $ andbysub stit u ti on  we  c a n e a si lyf ind  that $$ a- b= 3\ \\tex t { o r }\ \  a^ 2+b^ 2-2ab =9.$$
I got s tuc k at thi s poi nt...  how  t o get  the v aluefo r $ab$? Hints p leas e.

A:

S tar ti ngfr om $a - b=3$,whi chis odd,  this i m pli es t h epositive primes ca nn o t  b e  bothodd. S o  o ne of themis  $2 $, t h e  othe r is an  odd pri me. As $a -b >0$, it 's  $b$ w hi chisequal to $ 2$, so  $a=3+b= 5$, a n d $a^2+b^2=29$ . 

A:

Pure ob f us c a ti o n.
N oth ing about r oots or a lgeb r aort he su m of$a ^ 2+  b^2$ are relevant.  That $ -1$ i s a root simp ly means $ a ( - 1)^2 + b (-1) - 3=  0$ or in othe r wor ds $a - b= 3$ or $a  = b+ 3$.  
Wh at does m a t ter is t hat  $a ,b$  ar e po sitive primes a nd a ll  primes  ex cept $2 $ a reodd . I fthey were  both od dpr im es  th en $a -b$ woul dbean  ev en nu m ber. So on e of  t he m  is  even s o  o n e  ofth em  is$2$ . So t he o t her  is $2+ 3 = 5$.
S o $ a ^2 +  b ^2  = 29$. 

 Q:

(L'Étr an ger de Cam us ) T he usa g e  of "se"  with "laisser + faire" 
This q ues tionis o n "se
 destinationFolder

Called_in a_crontab every 5 minute:
*/5_* *_*_* /usr/bin/rsync_sourceFolder_destinationFolder

For options, permissions,_exlude of special_files or directories see_man rsync.

 Q:

If_$-1$is_a root for $ax^2+bx−3$, find $a^2+b^2$

Given: -1 is a root for $ax^2+bx-3$, with $a,b$_being_positive primes,_$x\in_\Bbb_R$.
Find: the numeric value for_$a^2+b^2$.

Background: question asked in an_entrance exam_(Colégio Militar 2005).
My attempt: the other root is_$3/a$_and by substitution_we can easily find that $$a-b=3\ \ \text{or}\ \_a^2+b^2-2ab=9.$$
I got stuck at this point..._how to get_the_value_for $ab$? Hints please.

A:

Starting_from $a-b=3$, which is odd, this_implies the positive primes cannot be_ both odd. So one of them_is $2$, the other is an_odd prime. As $a-b>0$, it's_$b$ which_is equal to $2$, so_$a=3+b=5$, and $a^2+b^2=29$.

A:

Pure_obfuscation.
Nothing about_roots or algebra_or the sum of $a^2 +_b^2$ are relevant._ That $-1$ is a root_simply_means $a(-1)^2 +_b(-1)-3_=_0$ or_in other words_$a_- b_=3$_or $a = b+3$.  
What_does_matter is that $a,b$ are positive primes_and all primes except_$2$_are odd. If they_were both odd primes then_$a -b$ would be an even_number. _So one_of them is even so one of them is $2$. _So the other is $2+3 =_5$.
So $a^2 + b^2_= 29$.

_Q:

(L'Étranger_de Camus) The_usage_of "se"_with "laisser + faire"

This question is on_"se