krishnasty wrote:

Hi Guys,

came across this problem without a OA. Please help me find a solution:

x^2 - |x| - 12

--------------- >= 2x

x-3

Please post your solutions along with explanation.

---------------------------------------------

Consider KUDOS if you like my post!!!

I am assuming your question is:

\frac{{x^2 - |x| - 12}}{{x - 3}} >= 2xFirst of all, remember that we cannot cross multiply in this case. The reason is that we do not know the sign of (x - 3). To handle the 2x on the right hand side, bring it to the left hand side to get:

\frac{{x^2 - |x| - 12}}{{x - 3}} - 2x >= 0\frac{{-x^2 - |x| + 6x - 12}}{{x - 3}} >= 0Now take 2 cases.

Case 1: If x >= 0, |x| = x

\frac{{-x^2 - x + 6x - 12}}{{x - 3}} >= 0\frac{{x^2 - 5x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

Now,

x^2 - 5x + 12 has no real roots. It will always be positive. (because

b^2 - 4ac is negative so we know that it has no real roots. Put x = 0, you get 12 which is positive. So this quadratic will never touch the x axis and will always remain above it.)

So for the left hand side to be negative, x < 3. (x cannot be 3 because in that case the denominator will be 0.)

Since x has to be greater than 0, the only solution here is 0 <= x < 3.

Case 2: If x <= 0, |x| = -x

\frac{{-x^2 + x + 6x - 12}}{{x - 3}} >= 0\frac{{x^2 - 7x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

\frac{{(x - 4)(x - 3)}}{{x - 3}} <= 0or x <= 4

Since x has to be less than(or equal to) 0, the only solution here is x <= 0.

Combining the solutions above, we get x < 3 as the final solution.

(I have assumed that you are comfortable with quadratic expressions and solving for inequalities once you have factors. If you are not, let me know and I will point you to the related theory.)

+1 to you .