MATHS confusion.....

[hassam]

k^2 - 16 <0
what are the possible values of k?
this is what i did....

(k+4)(k-4) < 0
k+4 < 0 and k-4 < 0
bt this gives me a silly anser that k i less than 4 and k is also less than -4 .....


can anybody help me out to remove this confusion

 

[Xenon]

(k+4)(k-4) < 0 is not incorrect.
16 is equal to square of 4.
Using a^2 - b^2 = (a+b)(a-b), k^2 -16 = (k+4)(k-4)

Proof:
(k+4)(k-4)
=k(k-4) + 4(k-4)
=k^2 -4k +4k -16
= k^2 -16

Following from....
(k+4)(k-4) < 0
Solutions of k are -4 and 4
Since this is less than zero, the range of values covered is from -4 to 4
therefore the solution is -4<k<4

k>4 and k<-4 are incorrect range. Substituting , e.g, k=5 in the original eqn ( k^2 -16 < 0) gives 9 which is greater than zero.
Again, substituting ,e.g, k= -5 gives 9 which is greater than zero as well.

 

[OakMoon!]

This is an inequality. Since (k+4)(k-4)<0 you need to use a number line method or need to make a rough sketch of the Quadratic Curve in order to find the final answer. That's the basic concept of solving Quadratic inequalties.
In this question you have to state the values of x when the curve is negative and below the x axis. And that lies in the region -4<x<4.

Hope this helped

 

[hassam]

ya u r ryt ....hamid ali studied this topic 2day.....

 

[OakMoon!]

I didn't get you hassam. What do you mean by me studying the topic today? I studied it about 2 years back in Olevels Addmaths.