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Question

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


Solution

(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.

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