Question

find the zeros of the polynomial x square - 3 and verify the relationship between the zeros and the coefficients​

Answer 1

Solution :

Given polynomial is x2−3 

$\sf{</p><p></p><p>Here, a=1,b=0  and c=−3</p><p></p><p>}$

$\sf \red{x²−3=(x− 3 )(x− 3)}$

$\sf{So, the \: value \: of x² −3 is \: zero \: when \: x= 3 or x − 3}$

$\sf{Now, \: sum \: of \: zeroes =3−3=0=}$

$\sf \frac{0}{1} \: = \frac{ - b}{a}$

$\sf {Product  of \: zeroes =(3)(−3)=−3=}$

$\sf \frac{- 3}{1} = \frac{c}{a}$

$\sf \red{Hence \: verified.}$

Answer 2

Answer:

Solution :

Given polynomial is x2−3

01=−ba\sf \frac{0}{1} \: = \frac{ - b}{a}

1

0

=

a

−b

−31=ca\sf \frac{- 3}{1} = \frac{c}{a}

1

−3

=

a

c

Henceverified.\sf \red{Hence \: verified.}Henceverified.