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

What numbers are zeros of g(x) = x^2 - 2x - 4? take your time if you need to!-example-1
User Stonean
by
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1 Answer

3 votes

Answer:


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
User David Kaftan
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