Question

If one of the zeroes of a quadratic polynomial f(x) = 14x2-42k2x-9 is negative of other.Find the value of k

Answer 1

Heya Friend!!

Here's ur ans

Given :-

• One zeros is negative of other

Let the Zeros be a
and ( - a )

f ( x ) = 14x² - 42k² x - 9

As we know

$sum \: of \: zeros \: = \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$a \: + ( - a) = \frac{42 {k}^{2} }{14}$

$0 = {3 {k}^{2} }$

0/3 = k²

0 = k

So , k will equal to 0

•Verification

When we put value of k and after factorising we will get one zeros negative of other

=> 14x² - 42k² x - 9 = 0

14x² - 9 = 0

( √14 x )² - ( 3 )² = 0

By using identity

[ a² - b² = ( a + b ) ( a - b ) ]

So,

( √14x + 3 ) ( √14x - 3 ) = 0

* ( √14x + 3 ) = 0

x = -3/√14

* ( √14x - 3 ) = 0

x = 3/√14

Hence we get one zeros negative of other!!

@Altaf

Answer 2

Answer:

Let the Zeros be n and -n

f ( x ) = 14x² - 42k² x - 9

α+β = -b/a = -(-42k²)/14

n+(-n) = 3k²

0 = 3k²

k² = 0/3

k² = 0

k = 0