If \[\alpha \] be the number of person who take junk food, \[\beta \] be the number of person who take food at home and \[\alpha \] and \[\beta \] be the zeroes of quadratic polynomial \[f\left( x \right) = {x^2} - 3x + 2\], then find a quadratic polynomial whose zeroes are \[\dfrac{1}{{2\alpha + \beta }}\] and \[\dfrac{1}{{2\beta + \alpha }}\], which way of taking food you prefer and why?

Hint: Here we need to find the quadratic polynomial whose zeroes are given. We will first the product and the sum of the zeroes of the first quadratic polynomial. Then we will use this relation between the roots and the given roots of the second quadratic polynomial to obtain the second polynomial.

Formula used:
Standard equation of the quadratic equation is \[g\left( x \right) = {x^2} - \left( {{\rm{sum\, of\, roots}}} \right)x + \left( {{\rm{product\, of\, roots}}} \right)\]

Complete step-by-step answer:
Here we need to find the quadratic polynomial whose zeroes are given.
It is given that:-
Number of person who take junk food \[ = \alpha \]
Number of person who take food at home \[ = \beta \]
The first quadratic equation is
\[f\left( x \right) = {x^2} - 3x + 2\]
Now, we will find the roots of the quadratic polynomial by factoring the given polynomial by splitting the middle term.
\[ \Rightarrow f\left( x \right) = {x^2} - 2x - x + 2\]
On further simplification, we get
\[ \Rightarrow f\left( x \right) = x\left( {x - 2} \right) - \left( {x - 2} \right)\]
Now, we will take common factors from the two terms.

We know that the sum of the roots of the quadratic equation is equal to the negative of the ratio of the coefficient of \[x\] to the coefficient of \[{x^2}\].
Therefore,
\[\alpha + \beta = - \dfrac{{\left( { - 3} \right)}}{1} = 3\] ………….. \[\left( 1 \right)\]
We know that the sum of the roots of the quadratic equation is equal to the ratio of the constant term to the coefficient of \[{x^2}\].
\[\alpha \cdot \beta = \dfrac{2}{1} = 2\] ……………… \[\left( 2 \right)\]
We have been given the roots of the second quadratic polynomial i.e. \[\dfrac{1}{{2\alpha + \beta }}\] and \[\dfrac{1}{{2\beta + \alpha }}\].
First we will find the sum of these zeroes of the second polynomial.
Sum of roots \[ = \dfrac{1}{{2\alpha + \beta }} + \dfrac{1}{{2\beta + \alpha }}\]
On further simplification, we get
\[ \Rightarrow \] Sum of roots \[ = \dfrac{{2\alpha + \beta + 2\beta + \alpha }}{{\left( {2\alpha + \beta } \right)\left( {2\beta + \alpha } \right)}}\]
On adding the terms, we get
\[ \Rightarrow \] Sum of roots \[ = \dfrac{{3\alpha + 3\beta }}{{2\alpha \times 2\beta + 2\alpha \times \alpha + \beta \times 2\beta + \beta \times \alpha }}\]
On further simplification, we get
\[ \Rightarrow \] Sum of roots \[ = \dfrac{{3\left( {\alpha + \beta } \right)}}{{5\alpha \beta + 2\left( {{\alpha ^2} + {\beta ^2}} \right)}}\]
We know the algebraic identity is \[{\left( {a + b} \right)^2} - 2ab = {a^2} + {b^2}\]
Using this identity in above equation, we get
\[ \Rightarrow \] Sum of roots \[ = \dfrac{{3\left( {\alpha + \beta } \right)}}{{5\alpha \beta + 2\left[ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right]}}\]
Now, we will substitute the values from equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] here.
\[ \Rightarrow \] Sum of roots \[ = \dfrac{{3 \times 3}}{{5 \times 2 + 2\left[ {{3^2} - 2 \times 2} \right]}}\]
On further simplifying the terms, we get
\[ \Rightarrow \] Sum of roots \[ = \dfrac{9}{{10 + 2 \times 5}} = \dfrac{9}{{20}}\]
Now, we will find the product of these zeroes of the second polynomial.
Product of roots \[ = \dfrac{1}{{2\alpha + \beta }} \times \dfrac{1}{{2\beta + \alpha }}\]
On further simplification, we get
\[ \Rightarrow \] Product of roots \[ = \dfrac{1}{{\left( {2\alpha + \beta } \right)\left( {2\beta + \alpha } \right)}}\]
On multiplying the terms, we get
\[ \Rightarrow \] Product of roots \[ = \dfrac{1}{{2\alpha \times 2\beta + 2\alpha \times \alpha + \beta \times 2\beta + \beta \times \alpha }}\]
On further simplification, we get
\[ \Rightarrow \] Product of roots \[ = \dfrac{1}{{5\alpha \beta + 2\left( {{\alpha ^2} + {\beta ^2}} \right)}}\]
We know the algebraic identity is \[{\left( {a + b} \right)^2} - 2ab = {a^2} + {b^2}\]
Using this identity here, we get
\[ \Rightarrow \] Product of roots \[ = \dfrac{1}{{5\alpha \beta + 2\left[ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right]}}\]
Now, we will substitute the values from equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] here.
\[ \Rightarrow \] Product of roots \[ = \dfrac{1}{{5 \times 2 + 2\left[ {{3^2} - 2 \times 2} \right]}}\]
On further simplifying the terms, we get
\[ \Rightarrow \] Product of roots \[ = \dfrac{1}{{10 + 2 \times 5}} = \dfrac{1}{{20}}\]
We know that the standard equation of the quadratic equation is given by
\[g\left( x \right) = {x^2} - \left( {{\rm{sum\, of\, roots}}} \right)x + \left( {{\rm{product\, of\, roots}}} \right)\]
Now, we will substitute the value of the sum and product of the roots here.
\[ \Rightarrow g\left( x \right) = {x^2} - \dfrac{9}{{20}} \times x + \dfrac{1}{{20}}\]
On further simplification, we get
\[ \Rightarrow g\left( x \right) = {x^2} - \dfrac{9}{{20}}x + \dfrac{1}{{20}}\]
This is the required quadratic polynomial.
I prefer to eat the food made at home because it is healthy for our body and it keeps us away from the harmful diseases.

Note: Here, roots of the quadratic equation are defined as the values of the variable of the quadratic equation which when substituted in the equation give the value zero. It is also called the solution of the equation. A quadratic equation is an equation which has a highest degree of variable as 2 and has 2 solutions. We can make a mistake by not writing the preference of food intake and end up giving half the solution of the question.