Question

What numbers are zeros of g(x) = x^2 - 2x - 4? take your time if you need to!

Answer 1

`x = 1 + \sqrt5, x = 1 - \sqrt5`

Explanation:

Hello!

We can solve the quadratic by using the quadratic formula.

Standard form of a quadratic:
`ax^2 + bx + c = 0`

Quadratic Formula:
`x = (-b\pm√(b^2 - 4ac))/(2a)`

Given our Equation:
`g(x) = x^2 - 2x - 4`

• a = 1
• b = -2
• c = -4

Plug the values into the equation and solve.

Solve

• 
  `x = (-b\pm√(b^2 - 4ac))/(2a)`
• 
  `x = (-(-2)\pm√((-2)^2 - 4(1)(-4)))/(2(1))`
• 
  `x = (2\pm√(4 +16))/(2)`
• 
  `x = (2\pm√(20))/(2)`
• 
  `x = (2\pm√(4 * 5))/(2)`
• 
  `x = (2\pm(\sqrt4 * \sqrt5))/(2)`
• 
  `x = (2\pm2\sqrt5)/(2)`
• 
  `x = 1 + \sqrt5, x = 1 - \sqrt5`