For three numbers x, y and z, x + y + z = 5, and xy + yz +

Question

For three numbers x, y and z, x + y + z = 5, and xy + yz + xz = 3. What is the largest value that x can take ? 

(A) 3*sqrt(13) 
(B) sqrt(19) 
(C) sqrt(17) 
(D) sqrt(15)

HSWKM · Answer

is it C ? (a+b+c)^2=a^2+b^2+c^2 +2ab+2ac+2bc 5^2=a^2+b^2+c^2+2(ab+bc+ac) 25=a^2+b^2+c^2 + 2(3) 19=a^2+b^2+c^2 If b, c are 1 each; a or x will be maximum a=sqrt(17) :?:

ssumitsh · Answer

Hi HSWKM

I agree with you till 19=a^2+b^2+c^2

but I fail to understand why a will be maximum when b,c are each = 1.

Here is an example, take b= 0.0001 and c = 0.0001

so 19 = a^2 + 2* 10^(-8) i.e. a = sqrt (19 - 2* 10 ^ (-8)) which is very close to sqrt (19) and therefore greater than sqrt (17).

"(a+b+c)^2=a^2+b^2+c^2 +2ab+2ac+2bc 
5^2=a^2+b^2+c^2+2(ab+bc+ac) 
25=a^2+b^2+c^2 + 2(3) 
19=a^2+b^2+c^2 

If b, c are 1 each; a or x will be maximum 
a=sqrt(17)"

HSWKM · Answer

I presumed they are integers. 

If yes, answer would be sqrt(17) else sqrt (19). 

Having said this, will ETS let this sort of ambiguity to creep in.

Vijo · Answer

If x,y,z, are integers why can't y and z be 0 thus x =sqrt(19)?

HSWKM · Answer

Right! That is precisely what I am saying. The information in question is insufficient.

Andys · Answer

Hi

 

If you see problem again, following condition is also specified
xy + yz + xz = 3
So we can not consider y and z be 0

For three numbers x, y and z, x + y + z = 5, and xy + yz +

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