Are you real, rational, irrational or imaginary?

If p p and q q are odd integers , then the equation x 2 + 2 p x + 2 q = 0 x^2 + 2px + 2q = 0 cannot have __________ \text{\_\_\_\_\_\_\_\_\_\_} .

Imaginary roots Rational roots Irrational roots Real roots

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1 solution

Anurag Pandey
Aug 7, 2016

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

Using Sridharacharya formula :

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

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

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

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

( p + m ) ( p m ) = 2 q (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 :

E v e n ± E v e n = E v e n Even \pm Even = Even

O d d ± O d d = E v e n Odd \pm Odd = Even

And E v e n ± O d d = O d d Even \pm Odd = Odd .

Given p is odd so m also has to be odd as we know O d d × O d d 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 ) = 2 q (p+m)(p-m) = 2q

We get.

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

2 ( k + b + 1 ) ( k b ) = q \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 2 q p^2- 2q cannot be perfect square and so the roots of the quadratic equation cannot be rational .

The end.

😀😀

In the third line you have made a mistake.

it will be p 2 2 q p^2-2 q not p 2 4 q p^2-4 q

Kushal Bose - 4 years, 10 months ago

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Where??????? Error 404 not found .! Lol thanks a lot buddy . I have made subsequent changes.

Anurag Pandey - 4 years, 10 months ago

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Again there is an error below the line p 2 m 2 p^2-m^2

Kushal Bose - 4 years, 10 months ago

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@Kushal Bose Error 303 corrected. I hope now there is no mistake. Thanks again.

Anurag Pandey - 4 years, 10 months ago

Nice solution +1.didn't do it as rigorously but almost same👍👍

Racchit Jain - 4 years, 10 months ago

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