Question

If $\alpha \ and\ \beta $ are the zeros of the quadratic polynomial \[f\left( x \right) = {{x}^{2}}-p\left( x+1 \right)-c\], show that \[\left( \alpha +1 \right)\left( \beta +1 \right)=1-c\].

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\[f\left( x \right)={{x}^{2}}-p\left( x+1 \right)-c\] and we have to show that \[\left( \alpha +1 \right)\left( \beta +1 \right)=1-c\].
Now, to find this value we will be using the sum of roots and product of roots. We know that in a quadratic equation $a{{x}^{2}}+bx+c=0$.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$
Therefore, in \[f\left( x \right)={{x}^{2}}-p\left( x+1 \right)-c\]
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=-\left( -p \right)..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=-\left( p+c \right)...........\left( 2 \right) \\
\end{align}$
Now, we have to show that,
\[\left( \alpha +1 \right)\left( \beta +1 \right)=1-c\]
So, we will take LHS and prove it to be equal to RHS.
In LHS we have,
\[\begin{align}
  & \left( \alpha +1 \right)\left( \beta +1 \right)=\alpha \beta +\alpha +\beta +1 \\
 & =\alpha \beta +\left( \alpha +\beta \right)+1 \\
\end{align}\]
Now, we will substitute the value of $\alpha \beta \ and\ \left( \alpha +\beta \right)\ \ $ from (1) and (2). So, we have,
\[\begin{align}
  & \left( \alpha +1 \right)\left( \beta +1 \right)=-\left( p+c \right)+\left( -\left( -p \right) \right)+1 \\
 & =-p-c+p+1 \\
 & =1-c \\
\end{align}\]
Therefore, we have,
\[\left( \alpha +1 \right)\left( \beta +1 \right)=1-c\]
Now, since we have LHS = RHS.
Hence Proved.

Note: To solve these types of questions one must know how to find the relation between sum of roots, product of roots and coefficient of quadratic equation.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$