   ### Equations and Inequations - I

Problem : Find the smallest value of k for which both the roots of the equation  x2 - 8 k x+ 16 (k2 – k + 1) = 0 will be real roots ?
Solution : For both roots to be real first of all the Discriminant must be greater than zero
Hence,   ∆ ≥ 0
64k2-4 (16(k2-k+1) ) 0
64k2 -64k2 + 4k-4 0
k1
So the minimum value is k = 1

Problem : Find the maximum and minimum value of x such that x2 - 8 x + 6 < 0
Solution : We can first find where the expression x2 - 8 x + 6 = 0 holds true, thereon we can better decide which direction to go.
x2 - 8 x + 6 = 0 gives us,
x = -(-8) ± √64-4*6
x = 8 ± √64-16
x = 8 ± √48
x = 8 ± 4√3

So the minimum value of x is 8 – 4√3 and the maximum value is 8 + 4√3.
In between these two points the expression will be less than zero.

+ve ‹‹‹‹ +ve ‹‹‹ (8- 4√3) ›››››› -ve ‹‹‹‹ (8 + 4√3) ›››› +ve ›››› +ve

Problem: Find the values of x satisfying the following,
x2- 5x + 6 > 0, x2- 15x + 16 < 0 and x2 – 6x + 16 < 0
Solution: We shall solve each of the inequations seperately and then take the common values in each.
First equation, x2- 5x + 6 > 0
On solving you will get, x >2 & x < 3

Second equation, x2- 15x + 16 < 0
x = [15 ± √225-64 ] /2
x = [15 ± √161] /2
Thus we get,
x > [15 - √161] /2 & x <[15 + √161] /2

Third equation, x2 -6x - 16 < 0
On solving we get, x > - 2 & x < 8

The confusing but not difficult part now, taking the common values. We will better do it on a number line to avoid any confusion.

Hence from above it is very clear that the common region is 2 < x < 3

IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010