# Consider equation ax^2+bx+c=0 whose roots are x and y.Then find quadratic equation whose roots are x/y and y/x?

Jul 16, 2018

$a c {x}^{2} - \left({b}^{2} - 2 a c\right) x + a c = 0$.

#### Explanation:

Given that, $x \mathmr{and} y$ are the roots of the quadr. eqn. :

$a {x}^{2} + b x + c = 0$.

$\therefore x + y = - \frac{b}{a} , \mathmr{and} , x y = \frac{c}{a} \ldots \ldots \ldots \ldots \ldots \ldots . . \left(\ast\right)$.

Let, $X = \frac{x}{y} , \mathmr{and} , Y = \frac{y}{x}$.

Then, $X + Y = \frac{x}{y} + \frac{y}{x} = \frac{{x}^{2} + {y}^{2}}{x y}$,

$= \frac{1}{x y} \left\{{\left(x + y\right)}^{2} - 2 x y\right\}$,

$= {\left(x + y\right)}^{2} / \left(x y\right) - 2$,

$= \frac{{\left(- \frac{b}{a}\right)}^{2}}{\frac{c}{a}} - 2$.

$\therefore X + Y = \frac{{b}^{2} - 2 a c}{a c} \ldots \ldots \ldots \ldots \ldots . . \left({\ast}^{1}\right)$.

Also, $X \cdot Y = 1. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({\ast}^{2}\right)$.

Hence, the desired quadr. eqn. is given by,

${x}^{2} - \left(X + Y\right) x + X Y = 0 ,$

$i . e . , {x}^{2} - \left(\frac{{b}^{2} - 2 a c}{a c}\right) x + 1 = 0 ,$

$\mathmr{and} , a c {x}^{2} - \left({b}^{2} - 2 a c\right) x + a c = 0$.

Jul 16, 2018

The equation is $a c {x}^{2} - \left({b}^{2} - 2 a c\right) x + a c = 0$

#### Explanation:

Let the roots of the equation

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

be

$\alpha$ and $\beta$

Then,

$\alpha + \beta = - \frac{b}{a}$

and

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

The roots of the new equation are

$\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$

Then the sum of the roots are

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{{\alpha}^{2} + {\beta}^{2}}{\alpha \beta}$

$= \frac{{\left(\alpha + \beta\right)}^{2} - 2 \alpha \beta}{\alpha \beta}$

$= \frac{{\left(- \frac{b}{a}\right)}^{2} - 2 \cdot \frac{c}{a}}{\frac{c}{a}}$

$= \frac{{b}^{2} / {a}^{2} - 2 \frac{c}{a}}{\frac{c}{a}}$

$= \frac{{b}^{2} - 2 a c}{a c}$

The product of the roots is

$\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$

${x}^{2} - \left(\frac{{b}^{2} - 2 a c}{a c}\right) x + 1 = 0$
$a c {x}^{2} - \left({b}^{2} - 2 a c\right) x + a c = 0$