what are the roots of $f(x)=x^{2}-48$?\n-48 and 48\n-24 and 24\n$-8sqrt{3}$ and $8sqrt{3}$\n$-4sqrt{3}$ and…

what are the roots of $f(x)=x^{2}-48$?\n-48 and 48\n-24 and 24\n$-8sqrt{3}$ and $8sqrt{3}$\n$-4sqrt{3}$ and $4sqrt{3}$

what are the roots of $f(x)=x^{2}-48$?\n-48 and 48\n-24 and 24\n$-8sqrt{3}$ and $8sqrt{3}$\n$-4sqrt{3}$ and $4sqrt{3}$

Answer

Explanation:

Step1: Set the function equal to 0

$x^{2}-48 = 0$

Step2: Isolate the $x^{2}$ term

$x^{2}=48$

Step3: Take the square - root of both sides

$x=\pm\sqrt{48}$

Step4: Simplify the square - root

$\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}$, so $x = \pm4\sqrt{3}$

Answer:

D. $-4\sqrt{3}$ and $4\sqrt{3}$