Question

In quadratic equation, x square + bx +c = 0 if b and c are integer and Descriminant is perfect square, then prove that roots are integers​

Answer 1

Step-by-step explanation:

Given,

In quadratic equation,

${x}^{2} + bx +c = 0$

b and c are integer and Descriminant is perfect square.

To prove,

Roots are integers.

Proof,

For Quadratic equation,

${x}^{2}+bx+c=0$

$x = \frac{ -b ± \sqrt{ {b}^{2} - 4c} }{2}$

Now,

To prove the roots are integers,

we need to prove that Numerator is also even

Case I :

When b is even

$= > \sqrt{ {b}^{2} - 4c } \\ \\ is \: also \: even$

Therefore,

$( - b \: ± \sqrt{ {b}^{2} - 4c } ) \\ \\ is \: also \: even$

Case II :

When b is odd

$= > \sqrt{ {b}^{2} - 4c} \\ \\ is \: odd$

Therefore,

$( - b ± \sqrt{ {b}^{2} - 4c } ) \\ \\ is \: also \: even$

In both cases,

Numerator is even,

Hence,

The roots are integers.

Thus, Proved.