Question

If one zero of $2{{x}^{2}}-3x+k$ is reciprocal to the other, then find the value of $k$.

Answer 1

Hint: Here we know that the formula that if m and n are two roots of the quadratic equation of the form $a{{x}^{2}}+bx+c=0$, then the sum and product of the roots of m and n can be written as: $m+n=\dfrac{-b}{a}$ and $m.n=\dfrac{c}{a}$. We shall use these two relations to obtain the answer.

Complete step by step answer:
In this problem the polynomial is given as $2{{x}^{2}}-3x+k$. So, the corresponding quadratic equation is:
$2{{x}^{2}}-3x+k=0\ldots \ldots \ldots \left( i \right)$
We know that for a quadratic equation there will be two roots. Here one root is given as the reciprocal of the other root. We know that for a quadratic equation of the form $a{{x}^{2}}+bx+c=0$, the product of the roots is $\dfrac{c}{a}$, that is let m be two roots of the equation, then, $m.n=\dfrac{c}{a}$. Now consider equation (i), where $a=2,b=-3$ and $c=k$. It is also given that the roots are the reciprocal of the other. Now we have to find the value of $k$.
So, if we consider the one root as $t$, then the condition that the other root must be the reciprocal of this means that the other root will be $\dfrac{1}{t}$. And as we know that the product of the roots are,
$m.n=\dfrac{c}{a}$
So, here, $m=t,n=\dfrac{1}{t},a=2$.
So, we can write it as,
$\begin{align}
  & t.\dfrac{1}{t}=\dfrac{c}{2} \\
 & \Rightarrow c=1.2=k=2 \\
\end{align}$
So, the value of k is 2.

Note: Here an alternate method to find the value of p is by using the factors $\left( x-t \right)$ and $\left( x-\dfrac{1}{t} \right)$. That is, by multiplying then and equating them to zero, we will get the quadratic equation. From the equation you can find the value of p (second root).