Question

Find the quadratic polynomial when sum and product of the roots are $\dfrac{1}{4}$ and -1 respectively.

Answer 1

Hint:We solve this question by first assuming the roots of our equation as $\alpha $ and $\beta $. Then we get the values of \[~\alpha +\beta \] and \[~\alpha \beta \]. Then we find the quadratic equation with roots $\alpha $ and $\beta $ in terms of \[~\alpha +\beta \] and \[~\alpha \beta \]. Then we substitute the values of \[~\alpha +\beta \] and \[~\alpha \beta \], thereby we can find the required equation of the quadratic polynomial.

Complete step by step answer:
We are given that sum of the roots of a quadratic equation is $\dfrac{1}{4}$ and the product of the roots of the equation are -1.
Let us assume that the roots of our quadratic equation are $\alpha $ and $\beta $.
So, as we are given that the sum of the roots is $\dfrac{1}{4}$, we can write it as,
$\Rightarrow \alpha +\beta =\dfrac{1}{4}..........\left( 1 \right)$
As we are given that the product of the roots is -1, we get,
$\Rightarrow \alpha \beta =-1..........\left( 2 \right)$
Now let us consider the quadratic equation with roots $\alpha $ and $\beta $.
$\Rightarrow \left( x-\alpha \right)\left( x-\beta \right)$
We can write it by simplifying as,
$\begin{align}
  & \Rightarrow {{x}^{2}}-\alpha x-\beta x+\alpha \beta =0 \\
 & \Rightarrow {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0 \\
\end{align}$
Now let us substitute the values from equations (1) and (2). Then we get,
$\Rightarrow {{x}^{2}}-\left( \dfrac{1}{4} \right)x+\left( -1 \right)=0$
Now let us multiply the above equation with 4 on both sides. Then we get,
$\begin{align}
  & \Rightarrow 4\left( {{x}^{2}}-\left( \dfrac{1}{4} \right)x+\left( -1 \right) \right)=4\times 0 \\
 & \Rightarrow 4{{x}^{2}}-x-4=0 \\
\end{align}$
So, the quadratic equation with sum and product of roots $\dfrac{1}{4}$ and -1 respectively is $4{{x}^{2}}-x-4$.
Hence the answer is $4{{x}^{2}}-x-4$.

Note:
We can also solve this question in another method.
Let us consider the quadratic equation, \[~a{{x}^{2}}+bx+c=0\] with roots \[~\alpha \] and \[\beta \]. Then we have,
\[\begin{align}
  & \Rightarrow \alpha +\beta =\dfrac{-b}{a} \\
 & \Rightarrow \alpha \beta =\dfrac{c}{a} \\
\end{align}\]
As we are given that sum and the product of the roots are $\dfrac{1}{4}$ and -1, we get,
\[\begin{align}
  & \Rightarrow \dfrac{-b}{a}=\dfrac{1}{4} \\
 & \Rightarrow \dfrac{b}{a}=-\dfrac{1}{4}.........\left( 3 \right) \\
 & \Rightarrow \dfrac{c}{a}=-1..........\left( 4 \right) \\
\end{align}\]
Now consider the quadratic equation we have taken.
\[~a{{x}^{2}}+bx+c=0\]
Now, let us divide it with a. Then we get,
$\Rightarrow ~{{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=0$
Now let us substitute the values from equation (3) and (4). Then we get,
$\Rightarrow {{x}^{2}}-\left( \dfrac{1}{4} \right)x+\left( -1 \right)=0$
Now let us multiply the above equation with 4 on both sides. Then we get,
$\begin{align}
  & \Rightarrow 4\left( {{x}^{2}}-\left( \dfrac{1}{4} \right)x+\left( -1 \right) \right)=4\times 0 \\
 & \Rightarrow 4{{x}^{2}}-x-4=0 \\
\end{align}$
Hence the answer is $4{{x}^{2}}-x-4$.