Question

If $\alpha \ and\ \beta $are the zeros of the polynomial \[f\left( x \right)={{x}^{2}}-2x+3\], find a polynomial whose roots are $\alpha +2,\beta +2$.

Answer 1

Hint: We will be using the concept of quadratic equations to solve the problem. We will be using the concept of zeros of polynomials to find the sum of zeroes and product of zeroes also we will be using the method of representing a quadratic polynomial in terms of their roots.

Complete Step-by-Step solution:
Now, we have been given $\alpha \ and\ \beta $are the zeros of the polynomial\[f\left( x \right)={{x}^{2}}-2x+3\]. We have to find a polynomial whose roots are$\alpha +2,\beta +2$.
Now, we know that in a quadratic polynomial $a{{x}^{2}}+bx+c$.
$\begin{align}
  & \text{sum of zeros }=\dfrac{-b}{a} \\
 & \text{product of zeros }=\dfrac{c}{a} \\
\end{align}$
Therefore, in polynomial\[f\left( x \right)={{x}^{2}}-2x+3\]
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=2..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=3...........\left( 2 \right) \\
\end{align}$
Now, we know that a polynomial with $\alpha \ and\ \beta $ as roots can be represented as $k\left( {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \right)$ where k is any real number. Therefore, the polynomial with $\alpha +2,\beta +2$as root is,
$\begin{align}
  & =k\left( {{x}^{2}}-\left( \alpha +2+\beta +2 \right)x+\left( \alpha +2 \right)\left( \beta +2 \right) \right) \\
 & =k\left( {{x}^{2}}-\left( \alpha +\beta +4 \right)x+\left( \alpha \beta +2\alpha +2\beta +4 \right) \right) \\
 & =k\left( {{x}^{2}}-\left( \alpha +\beta +4 \right)x+\left( \alpha \beta +2\left( \alpha +\beta \right)+4 \right) \right) \\
\end{align}$
Now, we will substitute the value of $\alpha +\beta \ and\ \alpha \beta $ from (1) & (2).
$\begin{align}
  & k\left( {{x}^{2}}-\left( 2+4 \right)x+\left( 3+2\left( 2 \right)+4 \right) \right) \\
 & k\left( {{x}^{2}}-6x+\left( 3+4+4 \right) \right) \\
 & k\left( {{x}^{2}}-6x+\left( 11 \right) \right) \\
 & k\left( {{x}^{2}}-6x+11 \right) \\
\end{align}$
Since, the polynomial will give the same roots for any value of k. Therefore, we take k =1. The polynomial is $\left( {{x}^{2}}-6x+11 \right)$.

Note: To solve these type of questions one must know that a quadratic polynomial having $\alpha \ and\ \beta $ as roots can be represented as $k\left( {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \right)$.