Question

What are the roots of f(x) = x2 – 48?

Answer 1

`f(x) = 0\iff x^2-48=0 \iff x^2 = 48 \iff x=\pm √(48) = \pm 4√(3)`

Answer 2

we have

`f(x)=x^(2)-48`

To find the roots equate the function to zero

`x^(2)-48=0`

Step 1

Adds
`48` to both sides of the equation

`x^(2)-48+48=0+48`

`x^(2)=48`

Step 2

Take the square root of both sides

`\sqrt{x^(2)}= √(48)`

`x=(+/-)√(48) \\x1=+√(48)=4√(3) \\ x2=-√(48)=-4√(3)`

therefore

the answer is

The roots are

`x1=4√(3)`

`x2=-4√(3)`