Question

If $\alpha \ and\ \beta $ are the roots of ${{x}^{2}}+5x+4=0$, then the equation whose roots are $\dfrac{\alpha +2}{3}\ and\ \dfrac{\beta +2}{3}$, is
A. $9{{x}^{2}}+3x+2=0$
B. $9{{x}^{2}}-3x-2=0$
C. $9{{x}^{2}}+3x-2=0$
D. $9{{x}^{2}}-3x+2=0$

Answer 1

Hint: We will be using the concept of quadratic equation to solve the problem. We will be using the concepts of sum of zeroes and product of zeroes of a quadratic equation to further simplify the solution. We will be using the method of representing a quadratic polynomial with the help of its roots.

Complete step-by-step solution -
Now, we have been given that $\alpha \ and\ \beta $ are the roots of a quadratic equation ${{x}^{2}}+5x+4=0$.
Now, we know that if $a{{x}^{2}}+bx+c=0$ is a quadratic equation then,
$\begin{align}
  & \text{sum of zeroes = }\dfrac{-b}{a} \\
 & \text{product of zeroes = }\dfrac{c}{a} \\
\end{align}$
So, for \[f\left( x \right)={{x}^{2}}+5x+4=0\]. We have,
$\begin{align}
  & \alpha +\beta \ =\ \text{sum of zeroes = }-5.......\left( 1 \right) \\
 & \alpha \beta \ =\ \text{Product of zeroes = 4}........\left( 2 \right) \\
\end{align}$
Now, we know that a quadratic polynomial with $\alpha \ and\ \beta $ as roots can be represented as,
\[k\left( {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \right)\ where\ k\in R\].
Now, for $\dfrac{\alpha +2}{3}\ and\ \dfrac{\beta +2}{3}$ as roots the quadratic polynomial will be,
$\begin{align}
  & k\left( {{x}^{2}}-\left( \dfrac{\alpha +2}{3}+\dfrac{\beta +2}{3} \right)x+\dfrac{\left( \alpha +2 \right)\left( \beta +2 \right)}{3\times 3} \right) \\
 & =k\left( {{x}^{2}}-\left( \dfrac{\alpha +\beta +4}{3} \right)x+\dfrac{\alpha \beta +2\left( \alpha +\beta \right)+4}{3\times 3} \right) \\
 & =k\left( {{x}^{2}}-\left( \dfrac{\alpha +\beta +4}{3} \right)x+\dfrac{\alpha \beta +2\left( \alpha +\beta \right)+4}{3} \right) \\
\end{align}$
Now, we will substitute the value of $\alpha +\beta \ and\ \alpha \beta $ from (2) and (1). So, we have,
$\begin{align}
  & =k\left( {{x}^{2}}-\left( \dfrac{-5+4}{3} \right)x+\left( \dfrac{4+2\left( -5 \right)+4}{3\times 3} \right) \right) \\
 & =k\left( {{x}^{2}}+\dfrac{x}{3}+\dfrac{-2}{9} \right) \\
\end{align}$
Since, the polynomial is the same for any value of $k\in R$. So, we take k = 9.
The equation is $9{{x}^{2}}+3x-2=0$
Hence, the correct option is (B).

Note: To solve these types of questions it is important to know the concepts of sum of roots and product of roots. Also it is to be noted that a quadratic polynomial with $\alpha \ and\ \beta $ as roots can be written as,
\[k\left( {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \right)\ where\ k\in R\]