# CAT - Inequalities

Question:
Solve the inequality x3 – 5x2 + 8x – 4 > 0.
A.  (2, )
B.  (1, 2) U (2, )
C.  (-, 1) U (2, )
D.  (-, 1)

Explanation:

Let a, b, c be the roots of this cubic equation
a + b + c = 5
ab + bc + ca = 8
abc = 4

This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic equations}
The other approach is to use polynomial remainder theorem
If you notice, sum of the coefficients = 0
=>  P(1) = 0
=>  (x - 1) is a factor of the equation. Once we find one factor, we can find the other two by dividing the polynomial by (x - 1) and then factorizing the resulting quadratic equation.
(x - 1) (x - 2) (x - 2) > 0

Let us call the product (x - 1)(x - 2)(x - 2) as a black box.

If x is less than 1, the black box is a –ve number
If x is between 1 and 2, the black box is a +ve number
If x is greater than 2, the black box is a +ve number

Since we are searching for the regions where black box is a +ve number, the solution is as follows:
1 < x < 2 OR x > 2

Difficulty level 2
IIM CAT Preparation Tips: CAT - Inequalities

## Aug 1, 2013

### CAT - Inequalities

Question:
Solve the inequality x3 – 5x2 + 8x – 4 > 0.
A.  (2, )
B.  (1, 2) U (2, )
C.  (-, 1) U (2, )
D.  (-, 1)

Explanation:

Let a, b, c be the roots of this cubic equation
a + b + c = 5
ab + bc + ca = 8
abc = 4

This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic equations}
The other approach is to use polynomial remainder theorem
If you notice, sum of the coefficients = 0
=>  P(1) = 0
=>  (x - 1) is a factor of the equation. Once we find one factor, we can find the other two by dividing the polynomial by (x - 1) and then factorizing the resulting quadratic equation.
(x - 1) (x - 2) (x - 2) > 0

Let us call the product (x - 1)(x - 2)(x - 2) as a black box.

If x is less than 1, the black box is a –ve number
If x is between 1 and 2, the black box is a +ve number
If x is greater than 2, the black box is a +ve number

Since we are searching for the regions where black box is a +ve number, the solution is as follows:
1 < x < 2 OR x > 2