Question

find the zeroes of the polynomial of x2-3 and verify the relationship between the zeroes and cofficcients​

Answer 1

Given polynomial = x² - 3

Here,

1. a = 1
2. b = 0
3. c = -3

Solution :

${x}^{2} - 3 \: = \: (x \: - \: \sqrt{3} \: ) \: ( {x}^{2} \: - \: \sqrt{3} )$

━━━━━━━━━━━━━━━━━━━━━━━━━

So, the value of x² - 3 is zero when x = √3 or x = -√3

━━━━━━━━━━━━━━━━━━━━━━━━━

Threrfore , the zeroes of x² −3 are √3 and -√3

━━━━━━━━━━━━━━━━━━━━━━━━━

Now, sum of zeroes = √3 - √3 = 0 = 0\1 = -b\a

━━━━━━━━━━━━━━━━━━━━━━━━━

Product of zeroes = ( √3 )(-√3) = -3 = -3\1 = c\a

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

Given polynomial = x² - 3

Here,

a = 1

b = 0

c = -3

Solution :

{x}^{2} - 3 \: = \: (x \: - \: \sqrt{3} \: ) \: ( {x}^{2} \: - \: \sqrt{3} )x

2

−3=(x−

3

)(x

2

−

3

)

━━━━━━━━━━━━━━━━━━━━━━━━━

So, the value of x² - 3 is zero when x = √3 or x = -√3

━━━━━━━━━━━━━━━━━━━━━━━━━

Threrfore , the zeroes of x² −3 are √3 and -√3

━━━━━━━━━━━━━━━━━━━━━━━━━

Now, sum of zeroes = √3 - √3 = 0 = 0\1 = -b\a

━━━━━━━━━━━━━━━━━━━━━━━━━

Product of zeroes = ( √3 )(-√3) = -3 = -3\1 = c\a