Question

If one of the zeroes of the quadratic polynomial f(n) = 14x2 - 42k2x - 9 is negative of the other, find the value of k.

Answer 1

$\huge{\green{\underline{\red{\mathscr{Answer:-}}}}}$

• One zeros is negative of other

Let the Zeros be a

and ( - a )

f ( x ) = 14x² - 42k² x - 9

As we know

$sum \: of \: zero = \frac{cofficient \: of \: x}{cofficient \: of \: x {}^{2} } \\ a + ( - a) = \frac{42k {}^{2} }{14}$

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!!

$\huge{\orange{\underline{\purple{\mathscr{THANKS.}}}}}$