If and are odd integers , then the equation cannot have .
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Given equation: x 2 + 2 p x + 2 q = 0
Using Sridharacharya formula :
x = 2 − 2 p ± 4 p 2 − 8 q ⇒ x = − p ± p 2 − 2 q
Now the value p 2 − 2 q can takes positive , negative value . It can also take value which are not perfect square and maybe which are perfect square ( which we are gonna prove is not possible). You can put p = 5 and q= 1 and get value of p 2 − 2 q = 2 3 which is not a perfect square.
Now let's discuss about the possibility of p 2 − 2 q being a perfect square .
Let , p 2 − 2 q = m 2
p 2 − m 2 = 2 q
( p + m ) ( p − m ) = 2 q
Now since the right hand side is even number so left hand side should also be an even number . We now that :
E v e n ± E v e n = E v e n
O d d ± O d d = E v e n
And E v e n ± O d d = O d d .
Given p is odd so m also has to be odd as we know O d d × O d d is odd which we don't want and so m must be odd.
Now let's see terms on left. Let p = 2k + 1 and m = 2b + 1 ( since both are odd integers ). p + m = 2 ( k + b + 1) which is even and p - m = 2 (k - b ) which is even ( To be observed that k - b is also even ) So when we substitute these values in the equation : ( p + m ) ( p − m ) = 2 q
We get.
4 ( k + b + 1 ) ( k − b ) = 2 q
⇒ 2 ( k + b + 1 ) ( k − b ) = q
Right side is odd and left side is even so both cannot be equal. And hence , p 2 − 2 q cannot be perfect square and so the roots of the quadratic equation cannot be rational .
The end.
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