Question

If alpha and beta, are the roots of the equation 9x2
+ 6x + 1 = 0, then equation with roots

1 / alpha and 1 / beta
is

a) 2x^2
+ 3x + 8 = 0
b) x^2 + 6x + 9 = 0
c) x^2
+ 5x – 9 = 0
d) 2x^2
– 3x – 8 = 0​

Answer 1

Answer:

x²-6x+1

Step-by-step explanation:

Please see the attached image.

Answer 2

Step-by-step explanation:

$\alpha \: and \: \beta \: are \: the \: roots \\ \\ 9 {x}^{2} + 6x + 1 \\ \alpha + \beta = \frac{ - b}{a} \\ = > \alpha + \beta = \frac{ - 6}{9} \\ = > \alpha + \beta = \frac{ - 2}{3} \\ \\ and \: \: \: \alpha \beta = \frac{c}{a} \\ \: \: \alpha \beta = \frac{1}{9} \\ \\ \\ we \: have \: to \: find \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } \\ = \frac{ \alpha + \beta }{ \alpha \beta } \\ = \frac{ \frac{ - 2}{3} }{ \frac{1}{9} } \\ = \frac{ - 2 \times 9}{3} \\ = - 2 \times 3 = - 6$

$\alpha \: and \: \beta \: are \: the \: roots \\ \\ 9 {x}^{2} + 6x + 1 \\ \alpha + \beta = \frac{ - b}{a} \\ = > \alpha + \beta = \frac{ - 6}{9} \\ = > \alpha + \beta = \frac{ - 2}{3} \\ \\ and \: \: \: \alpha \beta = \frac{c}{a} \\ \: \: \alpha \beta = \frac{1}{9} \\ \\ \\ we \: have \: to \: find \: the \: equation \: whose \: roots \: are \: \frac{1}{ \alpha } \: \: and \: \\ \frac{1}{ \beta } \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } \\ = \frac{ \alpha + \beta }{ \alpha \beta } \\ = \frac{ \frac{ - 2}{3} }{ \frac{1}{9} } \\ = \frac{ - 2 \times 9}{3} \\ = - 6 \\ \\ \frac{1 }{ \alpha } \times \frac{1}{ \beta } \\ = \frac{1}{ \frac{1}{9} } \\ = 9 \\ \\ {x}^{2} + ( \frac{1}{ \alpha } + \frac{1}{ \beta } )x - \frac{1}{ \alpha } \times \frac{1}{ \beta } = 0 \\ = > {x}^{2} - 6x - 9 = 0 \: \: \: ans$