Answer
Chapter 6 - Section 6.2 - Exercise Set: 91 (Answer)
All positive values of ‘b’ are 9, 12 and 21.
Work Step by Step
Chapter 6 - Section 6.2 - Exercise Set: 91 (Solution)
Explanation :
$y^2 + by + 20$ = $(y + p)(y + q)$
Since $p\cdot q$ = +20,
1) p and q have to be of the same signs
For (p + q) = b and positive ‘b’ required,
2) both p and q cannot be of negative values, hence, only positive pairs are to be considered
For p = 1, q = 20, b = 21
For p = 2, q = 10, b = 12
For p = 4, q = 5, b = 9
There is no need to consider both positive signs further as the sum of ‘b’ will be same and cyclic again
So, all positive values of ‘b’ are 9, 12 and 21.
