Answer
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Hint:
We are asked to calculate the quadratic polynomial whose sum of the zero is 6 and the product of the zero is 4. So, we will take a quadratic polynomial and we will use the relation between zeroes and coefficient of the polynomial \[a{{x}^{2}}+bx+c=0\] to firm the required polynomial. We will use \[\alpha +\beta =\dfrac{-b}{a}\] and \[\alpha \beta =\dfrac{c}{a}\] to form the equation of the quadratic polynomial.
Complete step by step answer:
We are given that for any polynomial, \[\alpha \] and \[\beta \] are the roots (zeroes) of that polynomial, and also we are given that the sum of the zeroes is 6, i.e. \[\alpha +\beta =6\] and we have the product of the zero as 4, i.e. \[\alpha \beta =4.\] Now, we are asked to find the required polynomial which will satisfy the given condition. To find the polynomial, we should know about the relation between the roots (zeroes) and the quadratic polynomial.
So, we will learn about these relations and we know generally the quadratic polynomial is given as \[a{{x}^{2}}+bx+c=0\] where \[a\ne 0.\]
Now, we also know that if \[\alpha \] and \[\beta \] are the zero of \[a{{x}^{2}}+bx+c=0\] then the sum of roots will be
\[\text{Sum of roots}=\dfrac{-\text{coefficent of x}}{\text{coefficent }{{x}^{2}}}\]
\[\text{Product of roots}=\dfrac{\text{constant term}}{\text{coefficent }{{x}^{2}}}\]
Now, as the coefficient of x is b, the coefficient of \[{{x}^{2}}\] is a and the constant term is c, so we get,
\[\alpha +\beta =\dfrac{-b}{a}\]
\[\alpha \beta =\dfrac{c}{a}\]
Now as our equation is \[a{{x}^{2}}+bx+c=0\] so dividing each side by a, we get,
\[\dfrac{a}{a}{{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=\dfrac{0}{a}\]
Simplifying we get,
\[\Rightarrow {{x}^{2}}+\left( \dfrac{b}{a} \right)x+\dfrac{c}{a}=0\]
As \[\dfrac{-b}{a}=\alpha +\beta \] so \[\dfrac{b}{a}=-\left( \alpha +\beta \right)\] and \[\dfrac{c}{a}=\alpha \beta .\]
So, our equation can be written as
\[{{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0\]
Now, as we have \[\alpha +\beta =6\] and \[\alpha \beta =4.\] So, using them in the above equation we get the quadrilateral polynomial as
\[{{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0\]
\[\Rightarrow {{x}^{2}}-\left( 6 \right)x+4=0\]
On simplifying, we get,
\[\Rightarrow {{x}^{2}}-6x+4=0\]
So, our required quadratic equation is \[{{x}^{2}}-6x+4=0.\]
Note:
Students need to remember that we have \[\alpha +\beta =\dfrac{-b}{a},\] always take care of the negative sign else we will get the wrong solution. Also, remember that \[a\ne 0\] that’s why we were able to divide the equation by a. If it is not mentioned that a is not equal to 0 then we cannot divide the term by a.
We are asked to calculate the quadratic polynomial whose sum of the zero is 6 and the product of the zero is 4. So, we will take a quadratic polynomial and we will use the relation between zeroes and coefficient of the polynomial \[a{{x}^{2}}+bx+c=0\] to firm the required polynomial. We will use \[\alpha +\beta =\dfrac{-b}{a}\] and \[\alpha \beta =\dfrac{c}{a}\] to form the equation of the quadratic polynomial.
Complete step by step answer:
We are given that for any polynomial, \[\alpha \] and \[\beta \] are the roots (zeroes) of that polynomial, and also we are given that the sum of the zeroes is 6, i.e. \[\alpha +\beta =6\] and we have the product of the zero as 4, i.e. \[\alpha \beta =4.\] Now, we are asked to find the required polynomial which will satisfy the given condition. To find the polynomial, we should know about the relation between the roots (zeroes) and the quadratic polynomial.
So, we will learn about these relations and we know generally the quadratic polynomial is given as \[a{{x}^{2}}+bx+c=0\] where \[a\ne 0.\]
Now, we also know that if \[\alpha \] and \[\beta \] are the zero of \[a{{x}^{2}}+bx+c=0\] then the sum of roots will be
\[\text{Sum of roots}=\dfrac{-\text{coefficent of x}}{\text{coefficent }{{x}^{2}}}\]
\[\text{Product of roots}=\dfrac{\text{constant term}}{\text{coefficent }{{x}^{2}}}\]
Now, as the coefficient of x is b, the coefficient of \[{{x}^{2}}\] is a and the constant term is c, so we get,
\[\alpha +\beta =\dfrac{-b}{a}\]
\[\alpha \beta =\dfrac{c}{a}\]
Now as our equation is \[a{{x}^{2}}+bx+c=0\] so dividing each side by a, we get,
\[\dfrac{a}{a}{{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=\dfrac{0}{a}\]
Simplifying we get,
\[\Rightarrow {{x}^{2}}+\left( \dfrac{b}{a} \right)x+\dfrac{c}{a}=0\]
As \[\dfrac{-b}{a}=\alpha +\beta \] so \[\dfrac{b}{a}=-\left( \alpha +\beta \right)\] and \[\dfrac{c}{a}=\alpha \beta .\]
So, our equation can be written as
\[{{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0\]
Now, as we have \[\alpha +\beta =6\] and \[\alpha \beta =4.\] So, using them in the above equation we get the quadrilateral polynomial as
\[{{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0\]
\[\Rightarrow {{x}^{2}}-\left( 6 \right)x+4=0\]
On simplifying, we get,
\[\Rightarrow {{x}^{2}}-6x+4=0\]
So, our required quadratic equation is \[{{x}^{2}}-6x+4=0.\]
Note:
Students need to remember that we have \[\alpha +\beta =\dfrac{-b}{a},\] always take care of the negative sign else we will get the wrong solution. Also, remember that \[a\ne 0\] that’s why we were able to divide the equation by a. If it is not mentioned that a is not equal to 0 then we cannot divide the term by a.