If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ 2x\ +\ 3$, find a polynomial whose roots are $α\ +\ 2,\ β\ +\ 2$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ 2x\ +\ 3$.
To do:
We have to find the quadratic polynomial having $α+2$ and $β+2$ as its zeros.
Solution:
We know that,
The standard form of a quadratic equation is $ax^2+bx+c=0$, where a, b and c are
constants and $a≠0$
Comparing the given equation with the standard form of a quadratic equation, $a=1$, $b=-2$ and $c=3$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{-(-2)}{1}=2$.
Product of the roots $= αβ = \frac{c}{a} = \frac{3}{1}=3$.
Let the sum and product of the zeros of the given quadratic equation be $S$ and $P$ respectively.
Therefore,
$ \begin{array}{l}
S=( \alpha +2) +( \beta +2)\\
\\
\ \ \ =( \alpha +\beta ) +4\\
\\
\ \ \ =2+4\\
\\
\ \ \ =6\\
\\
P=( \alpha +2) \times ( \beta +2)\\
\\
\ \ =( \alpha \beta +2\alpha +2\beta +4)\\
\\
\ \ =( \alpha \beta +2( \alpha +\beta ) +4)\\
\\
\ \ =( 3+2( 2) +4)\\
\\
\ \ =( 3+4+4)\\
\\
\ \ =11
\end{array}$
The quadratic polynomial having the sum of the roots $S$ and product of the roots $P$ is $f(x)=k(x^2-(S)x+P)$, where $k$ is any non-zero real number.
Therefore,
$f(x)=k(x^2-(6)x+11)$
$f(x)=k(x^2-6x+11)$
The required quadratic polynomial is $f(x)=k(x^2-6x+11)$, where $k$ is any non-zero real number.
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