Question

If quadratic equation x^{2}+ax+\frac{b}{4}=0 has rational roots such that a and b are positive integers less than 6, then number of possible pairs of (a, b) is

a) 4
b) 5
c) 6
d) 7​

Answer 1

4 possible pairs of ( a , b) are  ( 1, 1)  , ( 2 , 3) , (2 , 4) , ( 3 , 5)  for quadratic equation x² + ax  + b/4 = 0 where a and b are positive integers less than 6

Given:

x² + ax  + b/4 = 0

Rational Roots

a and b are positive integers less than 6

To Find:

number of possible pairs of (a, b)

Solution:

Ax² + Bx + C = 0   has rational roots if Discriminant is perfect square where A, B and C are rational number

D = B² - 4AC  must be perfect Square

Step 1:

Find discriminant for x² + ax  + b/4 = 0

a² - 4(1)(b/4)

= a² -  b

Step 2:

Find all possible values of a and b for which  a² - b is perfect square

Table below showing row for value a , and column for value of b

and intersection of rows and column showing value of a² - b

b\a     1       2      3      4       5

1        0       3      8     15       24

2       -1       2      7     14       23

3       -2      1       6     13       22

4       -3       0     5     12       21

5       -4      -1      4     11       20

Pair of a and b are

( 1, 1)  , ( 2 , 3) , (2 , 4) , ( 3 , 5)

Hence 4 possible pairs of ( a , b)

Correct option is a) 4

( Although Question has missing RHS in Quadratic Equation but it should be 0 Hence x² + ax  + b/4 = 0)