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Keith_Richards [23]
2 years ago
15

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

Mathematics
1 answer:
ipn [44]2 years ago
7 0

Answer:

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

Step-by-step 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 = \frac{-b\pm\sqrt{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.

<h3>Solve</h3>
  • x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}
  • x = \frac{-(-2)\pm\sqrt{(-2)^2 - 4(1)(-4)}}{2(1)}
  • x = \frac{2\pm\sqrt{4 +16}}{2}
  • x = \frac{2\pm\sqrt{20}}{2}
  • x = \frac{2\pm\sqrt{4 * 5}}{2}
  • x = \frac{2\pm(\sqrt4 * \sqrt5)}{2}
  • x = \frac{2\pm2\sqrt5}{2}
  • x = 1 + \sqrt5, x = 1 - \sqrt5
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