if x is a positive integer x = ? (1) {(x-5)(x+4)}^1/2 >10

ok i see the mistake now and thanks a lot for pointing it out

i have reviewed the wiki guide (https://gmatclub.com/wiki/Algebra) on this and i see the following

consider the following example

x^2>x
x^2-x>0
x(x-1) >0
now either both x>0 and x-1>0 which translates to x>0 and x>1
or
both x<0 and x-1<0 which translates to x<0 and x<1
so the solution can be
x>1 or x<0

now coming back to the question
(x-5)(x+4) >100
therefore this means
(x-5)>100 or (x-5)<100 which translates to x>105 or x<105

(x+4) >100 or (x+4)<100 which translates to x>96 or x<96
combining both statements together we get the solution
as
x>96 or x<96

similarly
now coming back to the question
(x-5)(x+4) <121
therefore this means
(x-5)<121 or (x-5)>121 which translates to x<126 or x>126

so statement 1 does not help

(x+4) <121 or (x+4)>121 which translates to x<117 or x>125
combining both statements together we get the solution
as
x<117 or x>125
so statement 2 does not help
combining 2 statements together we get
x<117 or x>125

from this again it can be any value greater than 125 or less than 117

so shdnt E be the answer

wud appreciate u r help , if u can tell me where exactly i am going wrong
Many thanks

If x is a positive integer x = ?

(1) {(x-5)(x+4)}^1/2 >10
(2) {(x-5)(x+4)}^1/2 <11


Together, we see that 100 < (x - 5)(x + 4) < 121, where x is a positive integer

Note that x - 5 and x + 4 differ by 9

15 x 6 < 100
100 < 16 x 7 < 121
17 x 8 > 121

Thus x + 4 must be 16, i.e. x = 12

Hello again, vdhawan1!



If we have zero on the right side, this reasoning would be correct. Otherwise, it is not applicable. So, in this case, in order to apply this logic, we should transform the equation so that it will have zero.

Thus, we may write:
(x-5)(x+4) = x^2-x-20 and thus we obtain
x^2-x-20>100 and consequently
x^2-x-120>0


This last one we could have solved by factorizing (theoretically), as you suggested, only it is not practical in the given example, because the roots of this quadratic equation are not integers, and we are looking for an integer x. Moreover, from the form of the last equation, it is clear that we have infinite possibilities for x, so clearly it is not suff.

That’s why it’s better to left the equation in form:
x^2-x>120 =>
x(x-1)>120

and look at the second one.

The equation from the second condition, written as x^2-x-141<0, again, leaves more than one possibility for x and it’s impractical to solve the q. equation, again.

So, let’s leave it in the form x(x-1)<141

And, finally, we have
120< x(x-1)< 141

and at this stage it’s time to pick numbers . Luckily, only one possibility for x, namely x=12, satisfies the equation.

***
So, in general, if you have an eq. of the kind (x-a)(x-b)>c (or < c), you may apply the reasoning you described only if c=0. Otherwise, it won’t work, and we need to search for another approach.
P.S. I've PM you.

Greenoak

Buddy cant thank u enough for this help

now i see where i m going wrong

Kudos to u for this