 QUESTION

# If the zeroes of the polynomial $5{x^2} - 7x + k$ are reciprocal of each other, then find the value of $'k'$ .

Hint- In this question we will just suppose one zero or root to be say x then the other one being its reciprocal will be $\dfrac{1}{x}$ . We will take their product and equate their product to the general product of roots for a quadratic polynomial for $a{x^2} + bx + c$ product is $\dfrac{c}{a}$ .

Given polynomial is $5{x^2} - 7x + k$
To get the zeroes firstly equate this to 0
$\Rightarrow 5{x^2} - 7x + k = 0$
Let the zeroes be a and b.
Since they are reciprocal of each other so $a = \dfrac{1}{b}$ --(1)
(Using (1))
Product of zeroes $= a \times b = \dfrac{1}{b} \times b = 1$ ---(2)
We know that general product of zeroes $= \dfrac{c}{a}$
In our equation, on comparing it with general equation which is $a{x^2} + bx + c = 0$ we get,
a = 5, b = -7, c = k
Now, product of zeroes $= \dfrac{c}{a} = \dfrac{k}{5}$
On comparing this with (2) we get,
$\Rightarrow \dfrac{k}{5} = 1$
$\Rightarrow k = 5$
Hence, the required value of k is 5

Note- For such questions, just keep in mind that the zero of a polynomial is the value of X that causes the polynomial to = 0 (or make Y=0). Also, the product of zeroes of a polynomial $a{x^2} + bx + c$ is $\dfrac{c}{a}$ and sum of zeroes of polynomial is negative of the coefficient of $x$ by the coefficient of ${x^2}$ i.e $\dfrac{-b}{a}$