Question

Find the zeroes of the polynomial X²-3 and verify the relationship between the zeroes and coefficients?

Answer 1

Step-by-step explanation:

${x}^{2} - 3 = 0$

${x}^{2} = 3$

$x = \sqrt{3} \: \: or \: \: - \sqrt{3}$

let them be alpha and beta

$\alpha + \beta = 0 = - \frac{b}{a}$

$\alpha \beta = - 3 = \frac{c}{a}$

Answer 2

↦Solution : -

Using Identity : -

a² - b² =(a - b) (a + b)

By using it , we can write : -

x² - 3 = (x - √3) (x + √3)

So : -

The value of x² - 3 is 0 when x = √3 or -√3

Therefore , the zeroes of x² - 3 are √3 and -√3 ..

Now : -

Sum of zeroes = √3 -√3 = 0

= -(Cofficient of zeroes)/Cofficient of x²

Product of zeroes = -3 = -3/1

= Constant term/Cofficient of x²