Question

zeroes of p(x) =z²- 27 are___ and ___ .​

Answer 1

Answer:

$z ^{2} - 27 = 0 \\ \\ = > z ^{2} = 27 \\ \\ = > z = + \: or \: - \: \sqrt{27} \\ \\ = > z = + 3 \sqrt{3} \: or \: - 3 \sqrt{3}$

⭐ zeroes oF p (x) = z^2 - 27 aRe +3√3 and -3√3 .

hope out helps uH ✔

Answer 2

Answer:

$z^2 - 27 = 0[a^2 - b^2 = (a + b)(a - b)\\\\0 = (z + 9\sqrt3)(z - 9\sqrt3)\\\\\therefore z + 9\sqrt3 = \frac{0}{z-9\sqrt3}\\\\\therefore z + 9\sqrt3 = 0\\\\\therefore z = -9\sqrt3$

OR

$z^2 - 27 = 0[a^2 - b^2 = (a + b)(a - b)\\\\0 = (z + 9\sqrt3)(z - 9\sqrt3)\\\\\therefore z - 9\sqrt3 = \frac{0}{z + 9\sqrt3}\\\\\therefore z - 9\sqrt3 = 0\\\\\therefore z = 9\sqrt3\\\\\\\\Thus \ the \ zeroes \ of \ given \ p(x) = z^2 - 27 \ are \ (9\sqrt3) \ and \ (-9\sqrt3)$

Step-by-step explanation: