Question

If $\alpha $ and $\beta $ are zeros of the polynomial ${{x}^{2}}-a\left( x+1 \right)-b$ such that $\left( \alpha +1 \right)\left( \beta +1 \right)=0$ , find the value of b.

Answer 1

Hint: We have a polynomial given as: ${{x}^{2}}-a\left( x+1 \right)-b$. The roots of the polynomial are $\alpha $ and $\beta $ such that $\left( \alpha +1 \right)\left( \beta +1 \right)=0$. We need to find the values of $\alpha $ and $\beta $ first and then substitute in the equation $\left( \alpha +1 \right)\left( \beta +1 \right)=0$ to find the value of b.
To find the value of $\alpha $and $\beta $ use the identity for the sum of zeros and product of zeros of a polynomial. Since the polynomial given in the question is quadratic, so for a quadratic equation $a{{x}^{2}}+bx+c=0$;
Sum of zeros = $-\dfrac{b}{a}$
Product of zeros = $\dfrac{c}{a}$

Complete step by step answer:
We have a polynomial given as: ${{x}^{2}}-a\left( x+1 \right)-b$ whose are $\alpha $and $\beta $.
Rearranging the equation, we can write: ${{x}^{2}}-ax-a-b$
Since the polynomial is quadratic, so by using the identities:
Sum of zeros = $-\dfrac{b}{a}$
Product of zeros = $\dfrac{c}{a}$
For the given polynomial ${{x}^{2}}-a\left( x+1 \right)-b$, we can say:
Sum of zeros: $\alpha +\beta =-\dfrac{(-a)}{1}......(1)$
Product of zeros: $\alpha \beta =\dfrac{(-a-b)}{1}......(2)$
Now, we have an equation that gives the relation between $\alpha $and $\beta $as $\left( \alpha +1 \right)\left( \beta +1 \right)=0$
Solving the above equation, we get:
$\begin{align}
  & \Rightarrow \left( \alpha +1 \right)\left( \beta +1 \right)=0 \\
 & \Rightarrow \alpha \beta +\alpha +\beta +1=0......(3) \\
\end{align}$
Substitute the values of $\left( \alpha \beta \right),\left( \alpha +\beta \right)$ from equation (1) and (2) in equation (3), we get:
$\begin{align}
  & \Rightarrow -a-b+a+1=0 \\
 & \Rightarrow b=1 \\
\end{align}$

Hence b equals to 1.

Note: While applying the identity for the sum of zeros and product of zeros, always take care of the negative sign in the sum of zeros. Some deliberately miss out on the use of negative signs in the formula and this gives you the wrong value. Also, it was given in the question, that the polynomial is quadratic. For a higher degree of the polynomial, the formula for the sum of zeros and product of zeros changes accordingly.