Solving an inequality with an unknown denominator (1 Viewer)

[danz90]

I know this is really easy for many of uz... but for some reason i never get full marks for this kind of question, and i've never exactly understood the method well.

I usually get a mark for saying tha x =/= SOMETHING.
But then i dont get the other mark coz i get the wrong inequality or forever.

Can somebody please explain the standard procedure for answering these types of questions

 

[RiSeAgAiNSt2538]

there is two methods.

first say the bottom is (x - 1). then let x be more than 1 so the bottom is positive. solve that. then let x be less than 1 and solve that. you will have 4 inequalities, the two you said let, and the 2 you found. 2 will cancel out and you have ur answer. just graph it.

the other is to make the bottom positive. take (x-1) again. multiply both sides by (x-1) so you have (x-1)2 which is always positive. then solve quadratic. graph parabola. then graph on number line with solution.

 

[tommykins]

回复: Solving an inequality with an unknown denominator

Take x+3/x-2 >= 2 for example.

x+3/x-2 > 2

(x+3)(x-2) > 2(x-2)^2 ------ multiplying both sides by (x-2)^2 to get rid of x possibly being negative.

0 > 2(x-2)^2 - (x+3)(x-2) --- taking LHS to the RHS
0 > (x-2)[2(x-2) - (x+3)] ----- factorising
0 > (x-2)[2x-4 - x - 3]
0 > (x-2)[x-7] ---- we now then draw this parabola/graph this.

Since the domain is asking for where the graph is 0 > f(x), our answer is simply the domains where the graph is negative.

that is, 2 < x =< 7 and x =/= 2 as this makes the denominator = 0.

 

[RiSeAgAiNSt2538]

i prefer the method tommy explained much better than me btw.

 

[danz90]

Re: 回复: Solving an inequality with an unknown denominator

Take x+3/x-2 >= 2 for example.

x+3/x-2 > 2

(x+3)(x-2) > 2(x-2)^2 ------ multiplying both sides by (x-2)^2 to get rid of x possibly being negative.

0 > 2(x-2)^2 - (x+3)(x-2) --- taking LHS to the RHS
0 > (x-2)[2(x-2) - (x+3)] ----- factorising
0 > (x-2)[2x-4 - x - 3]
0 > (x-2)[x-7] ---- we now then draw this parabola/graph this.

Since the domain is asking for where the graph is 0 > f(x), our answer is simply the domains where the graph is negative.

that is, 2 < x =< 7 and x =/= 2 as this makes the denominator = 0.

thanks heaps dude

 

[midifile]

there is two methods.

first say the bottom is (x - 1). then let x be more than 1 so the bottom is positive. solve that. then let x be less than 1 and solve that. you will have 4 inequalities, the two you said let, and the 2 you found. 2 will cancel out and you have ur answer. just graph it.

the other is to make the bottom positive. take (x-1) again. multiply both sides by (x-1) so you have (x-1)2 which is always positive. then solve quadratic. graph parabola. then graph on number line with solution.

There is also a third method.

Using tommys question:
x+3/x-2 >= 2

After you say that x=/=2
You let x+3/x-2 = 2 [changing the >= with an = sign]
then solve for x
x+3 = 2x-4
x=7

So youve got the points 2 and 7, and then you just test other points by subbing them into the equation to see if they are true
test x=0, -3/2 /> 2 so incorrect
test x=5 8/3 > 2 so true
test x=10 13/8 /> 2 so incorrect

Therefore the answer is between 2 and 7
So 2<x> < x <= 7</x>

 

[vds700]

there is two methods.

first say the bottom is (x - 1). then let x be more than 1 so the bottom is positive. solve that. then let x be less than 1 and solve that. you will have 4 inequalities, the two you said let, and the 2 you found. 2 will cancel out and you have ur answer. just graph it.

the other is to make the bottom positive. take (x-1) again. multiply both sides by (x-1) so you have (x-1)2 which is always positive. then solve quadratic. graph parabola. then graph on number line with solution.

Theres like 5 methods.

I use Tommy's way, probably the most staraightforward.

 

[I-Love-Jesus]

Yeah, I either use Tommy's way or I just graph it and solve graphically.

 

[Aznmichael92]

i use tommy's method as well but except i bring everything to lhs rather than rhs