If the zeroes of the polynomial $5{x^2} - 7x + k$ are reciprocal of each other, then find the value of $'k'$ .
ANSWER
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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 $\dfrac{c}{a}$ .
Complete step-by-step answer: 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 $\dfrac{-b}{a}$