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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