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

k≥1

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