Question

If alpha and beta are the roots of the polynomial x^2+17x+1 then find the equation whose roots are
$\frac{1}{ \alpha } and \frac{1}{ \beta }$

Answer 1

$<h2><font color = red > hey there$


$<font color = black > here we go....$



$x {}^{2} + 17x + 1$



$\alpha \: + \: \beta \: = - 17$



$\alpha \beta = 1$



sum of the root of the required equation is :-


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



solving it we get :-


$\frac{ \alpha + \beta }{ \alpha \beta } = \frac{ - 17}{1} = - 17$




Product of alpha and beta are :-


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





Hope it helps you

Thanks for asking.......



$<b > prabhudutt$

Answer 2

$\huge\boxed{\texttt{\fcolorbox{Red}{aqua}{Hey Mate!!!}}}$

$<b><i><font face=Copper black size=4 color=blue>$

Here is your answer
a = 1
b = 17
c = 1
If alpha and beta are the roots of the polynomial then

$\alpha + \beta = \frac{ - b}{a} \\ = - \frac{17}{1} \\ = - 17$
and
$\alpha \times \beta = \frac{c}{a} \\ = \frac{1}{1} \\ = 1$
If the roots are 1/alpha and 1/beta
so

Sum. of Roots is
$\frac{1}{ \alpha } + \frac{1}{ \beta } \\ = \frac{ \beta + \alpha }{\alpha \beta } \\ = \frac{ - 17}{1} \\ = - 17$
Now Product of roots is
$= \frac{1}{ \alpha } \times \frac{1}{ \beta } \\ = \frac{1}{ \alpha \beta } \\ = \frac{1}{1} \\ = 1$
Hence the equation is
$x {}^{2} - ( \frac{1}{ \alpha } + \frac{1}{ \beta } )x + \frac{1}{ \alpha \beta } \\ x {}^{2} - ( - 17)x + 1 = 0 \\ x {}^{2} + 17x + 1 = 0$
$\large{\red{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\underline{\underline{\underline{Hope\:it\: helps\: you}}}}}}}}}}}}}}}$