QUADRATIC INEQUALITIES -> HELP PLEASE!!

[switch]

Greetings

Could somebody please give me the answer for the following quadratic inequality , could you also explain how you arrived at the answer , will really appreciate

(X²+4)/(X-3) < 0

I need the answer urgently , thanks in advance

 

[XPFMember]

Assalamoalaikum!!

one tip...first solve the equation....by replacing < with = ....then finally when u arrive at the answer..put the sign accordingly!!


first u make a try and tell what difficulty u have...i'm kinda busy..so am not posting the steps..till then u try and if u want post ur attempt here..hope u dont mind..
gud luck

 

[MAVtKnmJ]

Are you sure that it's (X²+4) divided by (X-3) or is it (X²+4) multiplied by (X-3)

In either case the answer is X < 3

Let me know if you need help with the steps.

 

[ahmed t]

why?
i understand why ur anser is write but when u solve it algebraicaly
shoudnt it be
X²+4<0
which has no solutions

its either multiplied as MAVtKnMJ said
or its X²-4

 

[MAVtKnmJ]

Well X²+4 < 0 has a solution but with imaginary roots. So we will ignore that unless we are asked to give the imaginary roots.

Thus, (X-3) < 0 would be one of the solutions which simplifies to x < 3

 

[ahmed t]

how is X-3<0
i dont understand how you got that

 

[MAVtKnmJ]

Well I meant only if it was (X²+4) . (X-3) < 0 instead of (X²+4)/(X-3) < 0

I think it's a mistake on the posters part but you never know!

 

[usman]

(X²+4)/(X-3) < 0
Multiplying both sides of the inequality by (X-3)² which is +ve (>0) for sure, gives
(X²+4)*(X-3) < 0
Now, we can see that there is a product of two functions [i.e. (X²+4) and (X-3) respectively]
We can surely say that
(X²+4) is always +ve ∀x ε R
Therefore, the product can only be -ve (<0) when (X-3) is -ve (<0)
(X-3) < 0
X < 3 is the solution to the given inequality

Hope that helps

 

[usman]

(X²+4) is always +ve ∀X ε R (positive for all real X)
This is bcos (X²+4) has all values greater than or equal to 4(the value of it is 4 @ x = 0)
U cn visualize this as the graph of y = x² shifted 4 units above the x-axis, implying that the resulting graph (Y = X²+4) has all y values +ve

NB: +ve ---> positive
-ve ----> negative

 

[Bimal]

the process followed by usman is quite reliable.Go with it......