Question

If $\alpha \ and\ \beta $are the zeros of the quadratic polynomial such that \[\alpha +\beta =24\ and\ \alpha -\beta =8\], find the quadratic polynomial having $\alpha \ and\ \beta $ as its zeros.

Answer 1

Hint: We will be using the concept of quadratic equations to solve the problem. We will be using sum of roots and product of roots to further simplify the problem.

Complete Step-by-Step solution:
Now, we have been given $\alpha \ and\ \beta $are the zeros of the quadratic polynomial such that \[\alpha +\beta =24\ and\ \alpha -\beta =8\], and we have to find a polynomial whose roots are $\alpha \ and\ \beta $.
Now, we know that if $\alpha \ and\ \beta $ are the roots of a quadratic equation then the quadratic polynomial is ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta $
Now, we have been given that,
\[\begin{align}
  & \alpha +\beta =24.........\left( 1 \right) \\
 & \alpha -\beta =8.............\left( 2 \right) \\
\end{align}\]
Now adding (1) + (2), we have,
$\begin{align}
  & 2\alpha =32 \\
 & \alpha =16 \\
\end{align}$
We will substitute $\alpha =16$ in (1) to get $\beta $. Therefore,
$\begin{align}
  & \beta =24-\alpha \\
 & =24-16 \\
 & \beta =8 \\
\end{align}$
Now, we have to find the polynomial which has $\alpha =16,\beta =8$ as roots and we know that the polynomial will be in form as,
$\begin{align}
  & p\left( x \right)={{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \\
 & ={{x}^{2}}-\left( 16+8 \right)x+16\times 8 \\
 & ={{x}^{2}}-24x+128 \\
\end{align}$
Therefore, the polynomial is ${{x}^{2}}-24x+128$.

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