Question

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

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