The problem asks us to rewrite the quadratic expression $x^2 - 2x - 4$ in the form $(x + c)^2 + d$ and find the values of $c$ and $d$. Then, using this result, we are asked to solve the equation $x^2 - 2x - 4 = 0$, expressing the solution in the form $x = f \pm \sqrt{g}$, where $f$ and $g$ are integers.

1. Problem Description

The problem asks us to rewrite the quadratic expression $x^2 - 2x - 4$ in the form $(x + c)^2 + d$ and find the values of $c$ and $d$. Then, using this result, we are asked to solve the equation $x^2 - 2x - 4 = 0$, expressing the solution in the form $x = f \pm \sqrt{g}$, where $f$ and $g$ are integers.

2. Solution Steps

a) Completing the square for $x^2 - 2x - 4$:

We want to rewrite $x^2 - 2x - 4$ in the form $(x + c)^2 + d$.

Expanding $(x + c)^2 + d$, we get $x^2 + 2cx + c^2 + d$.

Comparing the coefficients of $x$ in $x^2 - 2x - 4$ and $x^2 + 2cx + c^2 + d$, we have $2c = -2$, so $c = -1$.

Then $(x - 1)^2 + d = x^2 - 2x + 1 + d$.

Since $x^2 - 2x + 1 + d = x^2 - 2x - 4$, we have $1 + d = -4$, so $d = -5$.

Therefore, $x^2 - 2x - 4 = (x - 1)^2 - 5$.

b) Solving $x^2 - 2x - 4 = 0$:

From part a), we know that $x^2 - 2x - 4 = (x - 1)^2 - 5$.

Thus, the equation $x^2 - 2x - 4 = 0$ can be rewritten as $(x - 1)^2 - 5 = 0$.

Adding 5 to both sides gives $(x - 1)^2 = 5$.

Taking the square root of both sides gives $x - 1 = \pm \sqrt{5}$.

Adding 1 to both sides gives $x = 1 \pm \sqrt{5}$.

3. Final Answer

a) $c = -1$, $d = -5$.

b) $x = 1 \pm \sqrt{5}$.