Question

If one zero of the quadratic polynomial (k2 + k)x2 + 68x + 6k is reciprocal of the other, find k

Answer 1

Hi Mate !!


Let one Zeros be a
then it's given that other zeros is reciprocal of first one
So, another Zeros is 1/a

[ reciprocal of a number is the no. which when multiplied by the first no. gives 1 as remainder ]


Given equation :- ( k² + k )x² + 68x + 6k


• Product of Zeros :-

$= \frac{constant \: \: term}{coefficient \: of \: {x}^{2} }$

$a \times \frac{1}{a} = \frac{6k}{ {k}^{2} + k}$


$1 = \frac{6k}{k(k + 1)}$



$1 = \frac{6}{k + 1}$


k + 1 = 6

k = 6 - 1

k = 5

So, the value of k is 5 !!

Answer 2

Answer:

Explanation:

Let one Zeros be a

then it's given that other zeros is reciprocal of first one

So, another Zeros is 1/a

[ reciprocal of a number is the no. which when multiplied by the first no. gives 1 as remainder ]

Given equation :- ( k² + k )x² + 68x + 6k

• Product of Zeros :-

k + 1 = 6

k = 6 - 1

k = 5

So, the value of k is 5