To answer your second question, The issue you run into is positive negative - you can't take a square root on both sides because you have to accomodate the negative option - in your example if x^2<4 then -2>X <2. This isn't quite the same for x^2>4 - because here EITHER X>2 or x<-2.

The same issue is going to come up with taking the square of each side of the inequalities you have listed - if there is a possiblity for the number to be negative (as with <11) you will have to switch the signs.

tpr-becky wrote:To answer your second question, The issue you run into is positive negative - you can't take a square root on both sides because you have to accomodate the negative option - in your example if x^2<4 then -2>X <2. This isn't quite the same for x^2>4 - because here EITHER X>2 or x<-2.

The same issue is going to come up with taking the square of each side of the inequalities you have listed - if there is a possiblity for the number to be negative (as with <11) you will have to switch the signs.

so how do i attack this DS

apoorva.srivastva wrote:What is the value of x, if x is an integer?

1.) [(x+4)*(x-5)]^1/2 < 11
2.) [(x+4)*(x-5)]^1/2 > 10

OA is [/spoiler]C[spoiler]

please tell
1.)whether we can square both the sides of an equality as in the question above??
2.) can we take square root on both the sides of inequalities eg. X^2 < 4 ??

Remember that if sqrt(k) < 11, 0 < k < 121

Thus (1) tells us that 0 < (x+4)(x - 5) < 121

(2) tells us that (x+4)(x - 5) > 100


For this product to be positive, the two terms must have the same sign i.e x < -4 or x > 5

It's helpful to see here that the distance between the two terms is 9
6(15) < 100 7(16) = 112 8(17) = 132

Thus x+4 and x - 5 could be 16 and 7 or - 7 and -16

10 < sqrt((12+4)(12 -5)) < 11, but also 10 < sqrt((-11+4)(-11-5)) < 11


It makes sense that if there the solutions come in pairs, as y=(x+4)(x- 5) is symmetric about x= 1/2


I get E and not C

6(15) < 100 7(16) = 112 8(17) = 132

Thus x+4 and x - 5 could be 16 and 7 or - 7 and -16

10 < sqrt((12+4)(12 -5)) < 11, but also 10 < sqrt((-11+4)(-11-5)) < 11


It makes sense that if there the solutions come in pairs, as y=(x+4)(x- 5) is symmetric about x= 1/2


I get E and not C

Hey Kevin, I cannot comprehend this. Could you give a more detailed explanation?

What is the value of x, if x is an integer?

1.) [(x+4)*(x-5)]^1/2 < 11
2.) [(x+4)*(x-5)]^1/2 > 10

Statement 1 and 2 together

100 < (x+4)(x-5) < 121

x ......... (x+4)(x-5)
11 ....... 15 * 6 = 90
12 ....... 16 * 7 = 112
13 ....... 17 * 8 = 136

-10 ...... -6 * -15 = 90
-11 ...... -7 * -16 = 112
-12 ...... -8 * -17 = 136

x = 12 or -11

Answer: E

apoorva.srivastva wrote:
please tell
1.)whether we can square both the sides of an equality as in the question above??
2.) can we take square root on both the sides of inequalities eg. X^2 < 4 ??

If you increase power both sides, no of root will increase and vise verse.
So, we should be carefull.



X^2 < 4

Above can be written as

-2 < X < 2

umm..how about this?

we have
1. sqrt[(x+4)(x-5)] < 11 => as kevin pointed out, 0 < (x+4)(x-5) < 121
2. sqrt[(x+4)(x-5)] > 10 => (x+4)(x-5) > 100

Cant figure out the value of x from any individual statement..combining the two,

121 > (x+4)(x-5) > 100

expanding,

121 > x^2 - 5x + 4x -20 > 100
adding 20 to all sides,

141 > x^2 - x > 120

now what i did was pick numbers.. consider 11 -> 121 - 11 = 110 ..nope we need a bigger one..
12 -> 144 - 12 = 132 ..it fits! 13 -> 169 - 13 = 156.. I think the difference between the square of a number and the number itself would go on increasing from now..

So C should be our ans..not sure if I wud've cracked it under timed conditions, though!

yupp..forgot to consider the negative end of the number line...(