Question

If one root of the polynomial $f(x) = 5{x^2} + 13x + k$ is reciprocal of the other, then the value of k is:
A.0
B.5
C.$\dfrac{1}{6}$
D.6

Answer 1

Hint: In this question, we need to determine the value of the constant term ‘k’ in the equation $f(x) = 5{x^2} + 13x + k$ such that the roots of the given equation are reciprocal to one another. For this, we will use the general property of the quadratic equation which is given as the product of the roots is the ratio of the constant term and the coefficient of ${x^2}$ , i.e., $\alpha \times \beta = \dfrac{c}{a}$

Complete step-by-step answer:
An algebraic equation with two as the degree is known as the Quadratic equation. Here, we have been given the equation $f(x) = 5{x^2} + 13x + k$ which has two as the degree and so the given equation is a quadratic equation. The standard quadratic equation is $f(x) = a{x^2} + bx + c$ with the roots $\alpha $ and $\beta $.
Let one of the roots of the given quadratic equation $f(x) = 5{x^2} + 13x + k$ be ‘a’ then, according to the question, the second root of the given quadratic equation $f(x) = 5{x^2} + 13x + k$ will be ‘1/a’.
Now, comparing the given $f(x) = 5{x^2} + 13x + k$ equation with the standard quadratic equation $f(x) = a{x^2} + bx + c$ to determine the values of a, b and c.
$
  5{x^2} + 13x + k = a{x^2} + bx + c \\
   \Rightarrow a = 5 \\
   \Rightarrow b = 13 \\
   \Rightarrow c = k \\
 $
Now, the product of the roots is given as the ratio of the constant term and the coefficient of ${x^2}$.
Substituting $\alpha = a,\beta = \dfrac{1}{a},c = k{\text{ and }}a = 5$ in the formula $\alpha \times \beta = \dfrac{c}{a}$ to determine the value of k.
$
  \alpha \times \beta = \dfrac{c}{a} \\
  a \times \dfrac{1}{a} = \dfrac{k}{5} \\
  k = 5 \times 1 \\
   = 5 \\
 $
Hence, the value of ‘k’ in the equation $f(x) = 5{x^2} + 13x + k$ is 5 such that the roots are reciprocal to one another.
So, the correct answer is “Option B”.

Note: The sum of the roots of the quadratic equation is given as the ratio of the negation of the coefficient of $x$ and the coefficient of ${x^2}$ while the product of the roots is given as the ratio of the constant term and the coefficient of ${x^2}$. Mathematically, $\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \times \beta = \dfrac{c}{a}$.