Question

$\alpha \& \frac{1}{\alpha }$ are zeros of polynomial $4{x^2} - 2x + \left( {K + 4} \right)$, find the value of K.

Answer 1

 Hint: Zeroes of a polynomial $f\left( x \right)$ are the values of $x$which when plugged in a polynomial $f\left( x \right)$ makes it zero.
$\therefore f\left( 0 \right) = 0$
The polynomial in which the maximum power of a variable (called as degree) is 2 is called Quadratic Polynomial. The general form of a quadratic polynomial is $a{x^2} + bx + c$, where $a,b,c \in R\& a \ne 0$. It means $a,b\& c$are real constants and the value of $a$ cannot be zero because if $a$ become zero then the polynomial becomes linear. A quadratic polynomial has two zeroes. Let its zeroes be $\alpha \& \beta $. Now the relation between zeroes $\left( {\alpha ,\beta } \right)$of quadratic polynomial and the coefficients $\left( {a,b,c} \right)$ of that quadratic polynomial is
$\begin{gathered}
  sum{\text{ }}of{\text{ }}zeroes{\text{ = }}\frac{{ - coefficient{\text{ }}of{\text{ }}x}}{{coefficient{\text{ }}of{\text{ }}{x^2}}} \\
  {\text{ }}\therefore \alpha + \beta = \frac{{ - b}}{a} \\
\end{gathered} $
$\begin{gathered}
  {\text{product }}of{\text{ }}zeroes{\text{ = }}\frac{{cons\tan t}}{{coefficient{\text{ }}of{\text{ }}{x^2}}} \\
  {\text{ }}\therefore \alpha \times \beta = \frac{c}{a} \\
\end{gathered} $
As we don’t know the value of $\alpha $, we have to eliminate $\alpha $ using the above relations and find the value of $K$.

Complete step-by-step answer:
The given polynomial is $f\left( x \right) = 4{x^2} - 2x + \left( {K + 4} \right)$.
It is in the general form of quadratic polynomial in $x$, where $a = 4,b = - 2,c = \left( {K + 4} \right)$.
The given zeroes of polynomial $f\left( x \right)$ are $\alpha \& \frac{1}{\alpha }$. We can see that one of the zero is reciprocal of the other.
Now from the first relation between zeroes and coefficients we have
$\begin{gathered}
  {\text{ }}\alpha + \beta = \frac{{ - b}}{a} \\
   \Rightarrow \alpha + \frac{1}{\alpha } = \frac{{ - \left( { - 2} \right)}}{4} \\
   \Rightarrow \alpha + \frac{1}{\alpha } = \frac{1}{2} \\
\end{gathered} $
As we see that $K$ is not present in the above equation. So, we cannot find $K$from the first relation. Hence, we go for second relation. From second relation between zeroes and coefficients we have
$\begin{gathered}
  {\text{ }}\alpha \times \beta = \frac{c}{a} \\
   \Rightarrow \alpha \times \frac{1}{\alpha } = \frac{{\left( {K + 4} \right)}}{4} \\
   \Rightarrow {\text{ }}1 = \frac{{\left( {K + 4} \right)}}{4} \\
   \Rightarrow {\text{ }}4 = K + 4 \\
   \Rightarrow {\text{ }}K = 0 \\
\end{gathered} $

Hence the value of $K$is 0.

Note: The quadratic polynomial will have two real zeros only when its Discriminant is greater than or equal to zero. Discriminant is $D = {b^2} - 4ac$.
Hence for real zeroes $\begin{gathered}
  D \geqslant 0 \\
  {b^2} - 4ac \geqslant 0 \\
\end{gathered} $