Inequality Question

[ephesus32]

Advertisement - Members don't see this ad

Can somebody explain this inequality problem: (algebraically)

What are all the values of x for which (x-2)(x+5) < 0?

a) 2< x < 5
b) -2 < x < 5
c) -5 < x < 2
d) x < -5
e) x > 2




The answer is C.

 

[Sublimation]

Can somebody explain this inequality problem: (algebraically)

What are all the values of x for which (x-2)(x+5) < 0?

a) 2< x < 5
b) -2 < x < 5
c) -5 < x < 2
d) x < -5
e) x > 2




The answer is C.

Treat it like a Quadratic form. its like this (x-2)<0 and (X+5)>0 Now solve for X in both. and oyu will end up with C.

 

[ephesus32]

Treat it like a Quadratic form. its like this (x-2)<0 and (X+5)>0 Now solve for X in both. and oyu will end up with C.

Yeah, I saw that but why do you switch the inequality sign for (x+5)>0 ??

 

[FutureInternist]

Can somebody explain this inequality problem: (algebraically)

What are all the values of x for which (x-2)(x+5) < 0?

a) 2< x < 5
b) -2 < x < 5
c) -5 < x < 2
d) x < -5
e) x > 2


The answer is C.

So the first step is obviously to solve for x which is pretty easy giving you x = -5 & 2 (which are your critical points). Now you have to choose some test points to see exactly what your solution set is.
So pick a point less than the least value you have (say -6), pick a point b/w your two critical points (0 is easiest) & then one bigger than your largest value (say 3)
Now plug each of the test points into the original eqn to see if the solution holds true.
So for -6 we get -8*-1 which is > 0 so all values less than -5 are out.
Plugging in 0, you get -2*5 = -10 (which is <0) hence all values b/w -5 & 2 are part of the answer
Lastly, plugging in 3 --> 1*8=8 (not < 0) so all values > 2 are out.

So final answer becomes -5<x<2

Note - There is no need to do the actual multiplication on problems like these....for example for test point -6, just realize that (-6-2) is a (-) & so is (-6+5) and multiplying two negatives gives a positive.