Bunuel wrote:
For how many integer values of x, is |x – 6| > |3x + 6|?
(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite
Hi
sobby,
The actual way would be...
Since both sides are positive, square both sides
\((x-6)^2>(3x+6)^2..........x^2-12x+36>9x^2+36x+36.........8x^2+48x<0\)
8x(x+6)<0.....
For 8x(x+6) to be NEGATIVE, one of 8x and x+6 will be negative and other positive..
If x is positive, 8x will be positive and x+6 will also be positive..
So x should be NEGATIVE..Then 8x will be negative, and thus x+6 should be positive ..
For that \(x+6>0.....X>-6....\) But x<0.
So range becomes \(0>x>-6\)...
Values are \(-1,-2,-3,-4,-5\)
5 value
C
Now take x^2+6x+9 = |x+3|^2 <9 .. after simplifying, we get -3 <x+3<3
5 integer values.