The problem asks us to find the solutions of three quadratic equations using Vieta's formulas. The quadratic equations are $x^2 - 4x + 3 = 0$, $x^2 + 6x + 8 = 0$, and $x^2 + 5x - 24 = 0$.

1. Problem Description

The problem asks us to find the solutions of three quadratic equations using Vieta's formulas. The quadratic equations are $x^2 - 4x + 3 = 0$, $x^2 + 6x + 8 = 0$, and $x^2 + 5x - 24 = 0$.

2. Solution Steps

Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. For a quadratic equation of the form $ax^2 + bx + c = 0$, let the roots be $x_1$ and $x_2$. Then Vieta's formulas state:

$x_1 + x_2 = -\frac{b}{a}$

$x_1x_2 = \frac{c}{a}$

a) $x^2 - 4x + 3 = 0$

Here, $a = 1$, $b = -4$, and $c = 3$.

$x_1 + x_2 = -\frac{-4}{1} = 4$

$x_1x_2 = \frac{3}{1} = 3$

We are looking for two numbers that add up to 4 and multiply to

3. These numbers are 1 and

3. Therefore, the solutions are $x_1 = 1$ and $x_2 = 3$.

b) $x^2 + 6x + 8 = 0$

Here, $a = 1$, $b = 6$, and $c = 8$.

$x_1 + x_2 = -\frac{6}{1} = -6$

$x_1x_2 = \frac{8}{1} = 8$

We are looking for two numbers that add up to -6 and multiply to

8. These numbers are -2 and -

4. Therefore, the solutions are $x_1 = -2$ and $x_2 = -4$.

c) $x^2 + 5x - 24 = 0$

Here, $a = 1$, $b = 5$, and $c = -24$.

$x_1 + x_2 = -\frac{5}{1} = -5$

$x_1x_2 = \frac{-24}{1} = -24$

We are looking for two numbers that add up to -5 and multiply to -

2

4. These numbers are 3 and -

8. Therefore, the solutions are $x_1 = 3$ and $x_2 = -8$.

3. Final Answer

a) $x_1 = 1, x_2 = 3$

b) $x_1 = -2, x_2 = -4$

c) $x_1 = 3, x_2 = -8$