Answer:
See explanation
Step-by-step explanation:
You forgot to add the quadratic equation, so we can't give you the exact answer you are looking for.
However this is a similar example:
Let the quadratic equation be:
[tex] {x}^{2} + 2x + 1 = 0[/tex]
If
[tex] \alpha \: \: and \: \: \beta[/tex]
are the zeros of this polynomial, we want to find a polynomial whose roots are
[tex] \frac{ \alpha^{2} }{ \beta} \: \: and \: \: \frac{ \beta^{2} }{ \alpha} [/tex]
The product of new roots is :
[tex]\frac{ \alpha^{2} }{ \beta} \times \frac{ \beta^{2} }{ \alpha} = \alpha \beta \\ = \frac{c}{a} \\ = \frac{1}{1} \\ = 1[/tex]
The sum of new roots is:
[tex]\frac{ \alpha^{2} }{ \beta} + \frac{ \beta^{2} }{ \alpha} = \frac{ { \alpha}^{3} + { \beta}^{3} }{ \alpha \beta} \\ = \frac{ { (\alpha + \beta)}^{3} - 3 \alpha \beta( \alpha + \beta) }{ \alpha \beta} \\ = \frac{( - { \frac{b}{a} )}^{3} - 3( \frac{c}{a} )( - \frac{b}{a} ) }{ \frac{c}{a} } \\ = \frac{( - { \frac{2}{1} )}^{3} - 3( \frac{1}{1} )( - \frac{2}{1} ) }{ \frac{1}{1} } \\ = \frac{8 + 6}{1} = 14[/tex]
The required polynomial is given by:
[tex] {x}^{2} - (sum \: of \: zeros) + sum \: of \: zeros = 0[/tex]
We substitute to get:
[tex] {x}^{2} - 8x + 1 = 0[/tex]
The new polynomial is
[tex]y = {x}^{2} - 8x + 1[/tex]
