Question

If one root of the polynomial $f(x)=5x^{2}+13x+k$ is reciprocal of the other, then the value of k is
(a) 0
(b) 5
(c) $\frac{1}{6}$
(d) 6

Answer 1

SOLUTION :

The correct option is (b) : 5.

Given : α  and 1/α are the zeroes of the  polynomial f(x) = 5x² + 13x + k

On comparing with ax² + bx + c,

a = 5, b= 13, c = k

Product of the zeroes = constant term/ Coefficient of x²

α × 1/α = c/a

1 = k/5

k = 5

Hence, the value of k is 5 .

HOPE THIS ANSWER WILL HELP YOU

Answer 2

Hi there!
Here's the answer:

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Given,

$f(x)=5x^{2}+13x+k$

The roots of this polynomial are reciprocal to each other.

Let one root be $\alpha$

So , the other root will be $\frac{1}{\alpha}$

We have,
Product of roots of polynomial $ax^{2}+bx+c$ is $\frac{c}{a}$

•°• Product of roots of given polynomial f(x) = $\frac{k}{5}$

$\alpha \frac{1}{\alpha}=\frac{k}{5}$

=> $1 = \frac{k}{5}$

=> $k = 5$

This answer k = 5 is in Option -b
Option - b is correct.

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