Given Information:

1. $f(x) = x^2 - 4x + 3$
2. $g(x) = 2x - 1$

We are solving the following tasks:
(i) Solve $f(x) = 0$
(ii) Solve $f(x) = g(x)$
(iii) Find $g(f(-2))$
(iv) Find $f(g(x))$

Solution:

(i) Solve $f(x) = 0$:

Set $f(x) = x^2 - 4x + 3 = 0$.
Factorizing the quadratic:
$x^2 - 4x + 3 = (x - 1)(x - 3) = 0$
So, $x = 1$ and $x = 3$.

(ii) Solve $f(x) = g(x)$:

Equate $f(x) = g(x)$:
$x^2 - 4x + 3 = 2x - 1$
Rearrange terms:
$x^2 - 6x + 4 = 0$
Factorizing $x^2 - 6x + 4$ using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \text{where } a = 1, b = -6, c = 4$ $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(4)}}{2(1)} = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm \sqrt{20}}{2} = \frac{6 \pm 2\sqrt{5}}{2}$ $x = 3 \pm \sqrt{5}$
Thus, $x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}$.

(iii) Find $g(f(-2))$:

First, calculate $f(-2)$:
$f(-2) = (-2)^2 - 4(-2) + 3 = 4 + 8 + 3 = 15$
Now substitute $f(-2) = 15$ into $g(x)$:
$g(15) = 2(15) - 1 = 30 - 1 = 29$
So, $g(f(-2)) = 29$.

(iv) Find $f(g(x))$:

Substitute $g(x) = 2x - 1$ into $f(x)$:
$f(g(x)) = f(2x - 1)$
$f(2x - 1) = (2x - 1)^2 - 4(2x - 1) + 3$
Expand $(2x - 1)^2$:
$(2x - 1)^2 = 4x^2 - 4x + 1$
Substitute into $f(g(x))$:
$f(g(x)) = 4x^2 - 4x + 1 - 8x + 4 + 3$
Simplify:
$f(g(x)) = 4x^2 - 12x + 8$
So, $f(g(x)) = 4x^2 - 12x + 8$.

Final Answers:

(i) $x = 1$, $x = 3$
(ii) $x = 3 + \sqrt{5}$, $x = 3 - \sqrt{5}$
(iii) $g(f(-2)) = 29$
(iv) $f(g(x)) = 4x^2 - 12x + 8$

Do you want details for any step?

Here are 5 related questions to deepen your understanding:

1. How do we determine the roots of a quadratic equation using factorization?
2. What is the quadratic formula, and when is it used?
3. How can we verify the solutions of $f(x) = g(x)$?
4. What happens if $g(x)$ is a higher-degree polynomial?
5. How do we interpret $f(g(x))$ in the context of composition of functions?

Tip: Always check your solutions by substituting them back into the original equations!