Inequalities / Quadratic Equation

Resurgent wrote:Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?

A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0

PS: I do not have the OA for this question.

Here's my solution. It includes a systematic way to solve quadratic inequalities.



The answer is not E since one solution to E is x=0 and x=0 is not a solution to the original inequality.

rohan_vus wrote:

Brent Hanneson wrote:

Resurgent wrote:Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?

A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0

PS: I do not have the OA for this question.

Here's my solution. It includes a systematic way to solve quadratic inequalities.



The answer is not E since one solution to E is x=0 and x=0 is not a solution to the original inequality.

With respect to the logic behind bold statement , one solution is x = -4 which satisfies option B but its not a solution to original inequality . This question seems kinda odd to me

In our answer choices, we are looking for an inequality such that every possible value for x from the original inequality [ (x-11)(x-3) < 0 ] must also be a solution to the inequality among the answer choices. You are ready the question the other way around.

x = -4 is not a solution to the original inequality (x-11)(x-3) < 0, so we need not consider that value for x when checking the answer choices. We need only consider values of x between 3 and 11.

When we plug these allowable values into the inquality in answer choice B, EVERY value of x is such that x^2 + 5x + 6 > 0

The same cannot be said for the other answer choices.

rohan_vus wrote:We need only consider values of x between 3 and 11

This is what i am driving at!.

All possible ranges are 3<x<11. So any of these values even satisfy option E.. So in what way option E is different than option B. Yo mentioned taking x = 0 for kicking out option E , but x = 0 is not in the range of the original inequality .
All values between 3 and 11 satisfy option E also .

Sorry, my bad. I was reading something completely different from what you were saying.
All possible values for x (3<x<11) make E valid as well.
Looks like the answer is B and E