Are you real, rational, irrational or imaginary?

Given equation: $x^2 + 2px + 2q =0$

Using Sridharacharya formula :

$x = \frac { -2p \pm \sqrt{4p^2 - 8 q } } {2}$ $\Rightarrow x = -p \pm \sqrt{p^2 - 2q}$

Now the value $p^2- 2q$ 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 $p^2 - 2q = 23$ which is not a perfect square.
Now let's discuss about the possibility of $p^2 - 2q$ being a perfect square .

Let , $p^2 - 2q = m^2$

$p^2 - m^2 = 2q$

$(p+m)(p-m)=2q$

Now since the right hand side is even number so left hand side should also be an even number . We now that :

$Even \pm Even = Even$

$Odd \pm Odd = Even$

And $Even \pm Odd = Odd$ .

Given p is odd so m also has to be odd as we know $Odd \times Odd$ 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 : $(p+m)(p-m) = 2q$

We get.

$4(k+b+1)(k-b) = 2q$

$\Rightarrow 2(k+b+1)(k-b)=q$

Right side is odd and left side is even so both cannot be equal. And hence , $p^2- 2q$ cannot be perfect square and so the roots of the quadratic equation cannot be rational .

The end.

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