Answer
Verified
370.5k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Onam is the main festival of which state A Karnataka class 7 social science CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Who was the founder of muslim league A Mohmmad ali class 10 social science CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers