Question

x square - 3
Find the zeroes of the following polynomial and find sum and product ​

Answer 1

Answer:

the zeroes are ✓3 and -✓3

hope it helps...

Answer 2

Let p(x) = ${x}^{2} - 3$

Zero of the polynomial is the value of x where,

p(x) = 0

Putting p(x) = 0

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

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

Using:-

${a}^{2} - {b}^{2} = (a - b)(a + b)$

$(x - \sqrt{3} )(x + \sqrt{3} ) = 0$

So,

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

Therefore,

$\alpha = \sqrt{3} \: and \: \beta = \sqrt{ - 3}$

are the zeros of the polynomial.

We can write,

$p(x) = {x}^{2} - 3$

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

$= {1x}^{2} + 0x - 3$

This equation is form of ${ax}^{2} + bx + c$

So,

$a = 1, \: b = 0, \: c = 3$