Question

Show that if x is real , the expression (x^2-bc)/(2x-b-c) has no real value between b and c.

Answer 1

(y-b)(y-c)>0

this statement is true for two conditions:-

1) when both (y-b) and (y-c) are positive then their products will be greater than 0

if both are positive then
y>b and y>c.
suppose, b is 4 and c is 6
then y should be greater than 6. so both (y-b) and (y-c) are positive.
For this case y cannot be 5 otherwise y-c will be negative.
so this shows that y can not be between b and c.

2) when both (y-b) and (y-c) are negative the their products will be greater than zero.
y<b and y< c

Lets take the same values b is 4 and c is 6.

then y should be less than 4 for both (y-b) and (y-c) to be negative.

For this case y should be less than 4.
In this case y cannot be 5 otherwise (y-b) for this case will be positive. and we want both the terms to be positive. so that their product is greater than 0.