1.

If α , β are the roots of ax² + bx + c = 0 where a , b , c belong to Rational numbers then prove thati] α + β is always rationalii] α - β is always irrationaliii] αβ is always rationaliv] α , β are conjugate to each other ​

Answer»

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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