Question

alpha and beta are the roots of the equation ax ^2 + bx + c=0 then the equation whose roots are alpha + 1/ beta and beta + 1/alpha is​

Answer 1

Step-by-step explanation:

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

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

So, Our needs are

$\alpha + \frac{1}{ \beta } \\ \beta + \frac{1}{ \alpha }$

Now, we want a equation with the sum -

$\alpha + \beta + \frac{1}{ \beta } + \frac{1}{ \alpha } \\ = \frac{ - b}{a} + \frac{ \alpha + \beta }{ \alpha \beta } \\ = \frac{ - b}{a} + \frac{\frac{ - b}{a}}{ \frac{c}{a} } \\ = \frac{ - b}{a } + \frac{ - b}{c} \\ = \frac{ - bc + - ab}{ac}$

We want the multiplication as -

$( \alpha + \frac{1}{ \beta } )( \beta + \frac{1}{ \alpha } ) \\ = \alpha \beta + 1 + 1 + \frac{1}{ \alpha \beta } \\ = \frac{c}{a} + 2 + \frac{a}{c} \\ = \frac{ {a}^{2} + {c}^{2} }{ac}+ 2 \\ = \frac{ {a}^{2} + {c}^{2} + 2ac }{ac} \\ = \frac{ {(a + c)}^{2}}{ac}$

The new formed equation is :

$p(x) = k( {x}^{2} + ( \frac{ {(a + c)}^{2}}{ac} + \frac{ - bc + - ab}{ac} )x + ( \frac{ {(a + c)}^{2}}{ac} \times \frac{ - bc + - ab}{ac}) \\ = ...$

you should now evaulate more to get the equation.