nishantsharma87 wrote:
anshumishra wrote:
15x - 2/x > 1
Case1 : x > 0
Multiply both sides by "x"
=> 15x^2 - 2 > x
=> 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5)
Since x > 0 , so x > 2/5
Case 2 : x < 0
Multiply both sides by "x"
=> 15x^2 - 2 < x
=> 15x^2 - x - 2 < 0
Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.
In the first case I got x> 2/5 and x> -1/3 and x>0 , which results in x>2/5 so no problem here.
In the second case I have a problem. I get x<2/5 and
x< -1/3 and x<0This results in x < -1/3
So my answer comes out to be x<-1/3 or x>2/5 which is option E , but the answer is D.
Can someone please explain what I am doing wrong here? Why should x be in the range -1/3<x<0 in the case when x<0 ?
A detailed explanation will be a great help because it seems something is wrong with my understanding of some fundamental inequality rules. A common mistkae that people do on inequalitites is to multiply by a variable when you dont know the sign. You do not have to confuse yourself with the 2 cases above. A straightforward solution is:
15x-2/x>1 ---> 15x-2/x-1>0 ---> \(\frac{15x^2-2-x}{x}>0\) ---> \(\frac{15x^2-x-2}{x}>0\) ---> \(\frac{15x^2-6x+5x-2}{x}>0\)---> \(\frac{(3x+1)(5x-2)}{x}>0\)
---> \(\frac{(x+1/3)(x-2/5)}{x}>0\) ....(1)
Now, you have 3 roots at x=-1/3, 0 and 2/5.
So if you draw a numberline as shown in the attached figure, you will see that the expression in (1) will be true for
Attachment:
10-21-15 9-51-48 AM.jpg [ 3.27 KiB | Viewed 1076 times ]
-1/3<x<0 and x>2/5, giving you D as the correct answer.
For your solution, the text in red is incorrect. After you get \(15x^2 - x - 2 <\)0, you can not get x<-1/3. It should be -1/3<x<0.
You can check it by assuming x<-1/3 or x=-2. You get \(15*(-2)^2-2-(-2) = 60-2+4 >0\) and NOT <0. Thus x<-1/3 DOES NOT satisfy the case 2 inequality of \(15x^2 - x - 2 < 0\).
Hope this helps.