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x^2+6x+p, p is constant ,what is p? 28 Dec 2005, 20:51
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x^2+6x+p, p is constant ,what is p?
1）x^2+6x+p=(x+r)^2 for all values of x, and for some value of r,
2) x^2+6x+p＝ (x-m)(x-n), for all values of x, and for some value of m,n.



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x^2+6x+p, p is constant ,what is p? 28 Dec 2005, 20:58
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"Practice makes perfect " so .... :P :

SQRT(8+√44)-SQRT(8-√44) = ?
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x^2+6x+p, p is constant ,what is p? 28 Dec 2005, 21:07
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:wink:

If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4
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x^2+6x+p, p is constant ,what is p? 28 Dec 2005, 21:48
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:wink:

If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4
Show more


A

(1) x = 4n + 2 ==> 3x + 1 ==> 12n + 7 ==> which is an odd number so the remainder will not be 0

(2) Insufficient. Tried x = 5, or 33....etc.
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x^2+6x+p, p is constant ,what is p? 28 Dec 2005, 22:31
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laxieqv
"Practice makes perfect " so .... :P :

SQRT(8+√44)-SQRT(8-√44) = ?
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:shock:
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 01:03
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"Practice makes perfect " so .... :P :

SQRT(8+√44)-SQRT(8-√44) = ?
Show more


I get 2Sqrt(Sqrt(11)). Here is my working.

SQRT(8+√44)-SQRT(8-√44) = SQRT(8+2√11)-SQRT(8-2√11)
= SQRT(2(4+√11))-SQRT(2(4-√11) )
= SQRT(2(4+√11-4+√11) )
= SQRT(4√11)
= 2SQRT(√11)
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 01:09
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laxieqv
:wink:

If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4
Show more


From stmt 1 we get x = 6,10,14,18,22,26... etc.
Substituting each value in 3x+1 and dividing by 10 leaves the remainder 9,1,3,5,7 respectively and the sequence of remainder continues to be 9,1,3,5,7 and is never zero. Hence the answer is no, and the statemtn is suff.

Stmt 2 let x = 5 then 3x+1 = 16 = 6 mod 10.
let x = 33 then 3x+1 = 100 = 0 mod 10 so the stmt is not suff.

Hence A.
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 01:14
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laxieqv
x^2+6x+p, p is constant ,what is p?
1）x^2+6x+p=(x+r)^2 for all values of x, and for some value of r,
2) x^2+6x+p＝ (x-m)(x-n), for all values of x, and for some value of m,n.
Show more


Took a while to understand. Here is my solution.

From stmt 1 we get
x^2+6x+p = x^2+2rx+r^2
which means 2rx = 6x r = 3 then (x+3)^2 becomes
x^2+6x+9, hence p=9.

So stmt 1 is suff.

From stmt 2 we get m+n=6, and we do not get the individual values of m and n. so insuff.
Hence A.
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 02:06
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Try this one as well :wink:

SQRT(12+√44)-SQRT(12-√44) = ?
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 02:08
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krisrini
laxieqv
"Practice makes perfect " so .... :P :

SQRT(8+√44)-SQRT(8-√44) = ?

I get 2Sqrt(Sqrt(11)). Here is my working.

SQRT(8+√44)-SQRT(8-√44) = SQRT(8+2√11)-SQRT(8-2√11)
= SQRT(2(4+√11))-SQRT(2(4-√11) )
= SQRT(2(4+√11-4+√11) )
= SQRT(4√11)
= 2SQRT(√11)
Show more


We can't do it this way coz squareroot is different from multiplying.
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 02:23
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laxieqv
krisrini
laxieqv
"Practice makes perfect " so .... :P :

SQRT(8+√44)-SQRT(8-√44) = ?

I get 2Sqrt(Sqrt(11)). Here is my working.

SQRT(8+√44)-SQRT(8-√44) = SQRT(8+2√11)-SQRT(8-2√11)
= SQRT(2(4+√11))-SQRT(2(4-√11) )
= SQRT(2(4+√11-4+√11) )
= SQRT(4√11)
= 2SQRT(√11)

We can't do it this way coz squareroot is different from multiplying.
Show more


Yes you are right. Let me assume A to be SQRT(8+√44)
and B to be SQRT(8-√44)
a-b = a^2-b^2/(a+b)
Using this I get 2 SQRT(44)/SQRT(2(4+√11))-SQRT(2(4-√11) )
= 4 SQRT(11)/(SQRT(2) *(SQRT(4+√11) - SQRT(4-+√11))
2√22/*(SQRT(4+√11) - SQRT(4-+√11)
Can't proceed any further...

Help....
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 20:33
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laxieqv
:wink:

If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4
Show more


OA is A, OE: refer to our friends' explanations :)
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 20:35
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krisrini
laxieqv
x^2+6x+p, p is constant ,what is p?
1）x^2+6x+p=(x+r)^2 for all values of x, and for some value of r,
2) x^2+6x+p＝ (x-m)(x-n), for all values of x, and for some value of m,n.

Took a while to understand. Here is my solution.

From stmt 1 we get
x^2+6x+p = x^2+2rx+r^2
which means 2rx = 6x r = 3 then (x+3)^2 becomes
x^2+6x+9, hence p=9.

So stmt 1 is suff.

From stmt 2 we get m+n=6, and we do not get the individual values of m and n. so insuff.
Hence A.
Show more


modelled solution, buddy~! Yup, OA is A.
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 20:45
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laxieqv
"Practice makes perfect " so .... :P :

SQRT(8+√44)-SQRT(8-√44) = ?
Show more



SQRT(8+√44)-SQRT(8-√44) = A
--> A^2= [ sqrt( 8+sqrt44) - sqrt(8- sqrt44)] ^2
= [sqrt(8+sqrt44)]^2 - 2*[sqrt(8+sqrt44)*sqrt( 8- sqrt44) + [sqrt(8-sqrt44)]^2
= 8+sqrt44 - 2*sqrt[(8+sqrt44)*(8-sqrt44)] + 8-sqrt44
= 16 - 2*sqrt[64- (sqrt44)^2]
= 16 - 2*sqrt(64-44)= 16- 2sqrt20
= 16- 4sqrt5

---> A= sqrt(16-4sqrt5)
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x^2+6x+p, p is constant ,what is p? 29 Dec 2005, 22:17
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laxieqv
"Practice makes perfect " so .... :P :

SQRT(8+√44)-SQRT(8-√44) = ?


SQRT(8+√44)-SQRT(8-√44) = A
--> A^2= [ sqrt( 8+sqrt44) - sqrt(8- sqrt44)] ^2
= [sqrt(8+sqrt44)]^2 - 2*[sqrt(8+sqrt44)*sqrt( 8- sqrt44) + [sqrt(8-sqrt44)]^2
= 8+sqrt44 - 2*sqrt[(8+sqrt44)*(8-sqrt44)] + 8-sqrt44
= 16 - 2*sqrt[64- (sqrt44)^2]
= 16 - 2*sqrt(64-44)= 16- 2sqrt20
= 16- 4sqrt5

---> A= sqrt(16-4sqrt5)
Show more


Reads very simple, after I read what you wrote. :-D



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