Question

$\huge {\mathfrak {\pink {\underline{ \underline{Question : - }}}}}$
If α , β are the roots of ax² + bx + c = 0 where a , b , c belong to Rational numbers then prove that

i] α + β is always rational

ii] α - β is always irrational

iii] αβ is always rational

iv] α , β are conjugate to each other

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Answer 1

SOLUTION

GIVEN

$α \: , β \: are \: the \: roots \: of \: ax² + bx + c = 0 \: where \: a , b , c \: \in \mathbb{Q}$

TO PROVE

i] α + β is always rational

ii] α - β is always irrational

iii] αβ is always rational

iv] α , β are conjugate to each other

CONCEPT TO BE IMPLEMENTED

A general equation of quadratic equation is

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

Now one of the way to solve this equation is by SRIDHAR ACHARYYA formula

For any quadratic equation

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

The roots are given by

$\displaystyle \: x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

EVALUATION

Sridhar Acharya Formula for finding the roots of the quadratic equation ax² + bx + c = 0

The roots of the quadratic equation

ax² + bx + c = 0 is given by

$\displaystyle \: x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

Let

$\displaystyle \: \alpha \: \: = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a}$

and

$\displaystyle \: \beta \: = \frac{ - b - \: \sqrt{ {b}^{2} - 4ac} }{2a}$

i)

$\displaystyle \: \alpha + \beta\: \:$

$\displaystyle \: = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} + \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$\displaystyle \: = - \frac{2b}{2a}$

$\displaystyle \: = - \frac{b}{a}$

Which is rational

So α + β is always rational

ii)

$\displaystyle \: \alpha - \beta\: \:$

$\displaystyle \: = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} - \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$\displaystyle \: = \frac{ 2 \sqrt{ {b}^{2} - 4ac} }{2a}$

$\displaystyle \: = \frac{ \sqrt{ {b}^{2} - 4ac} }{a}$

Which is irrational

So α - β is always irrational

iii)

$\displaystyle \: \alpha \beta\: \:$

$\displaystyle \: = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \times \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$\displaystyle \: = \frac{ {b}^{2} - ( {b}^{2} - 4 ac\: ) }{4 {a}^{2} }$

$\displaystyle \: = \frac{ {b}^{2} - {b}^{2} + 4 ac\: }{4 {a}^{2} }$

$\displaystyle \: = \frac{4ac}{4 {a}^{2} }$

$\displaystyle \: = \frac{c}{a}$

Which is rational

So αβ is always rational

iv)

Since α & β are of the form

$x + \sqrt{y} \: \: and \: \: x - \sqrt{y}$respectively

Also

$\displaystyle \: \alpha \beta \: = \frac{c}{a}$

So α , β are conjugate to each other

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Answer 2

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