It seems like the image contains a list of quadratic equations where you're asked to determine the sum and product of the roots, along with finding the actual roots for each equation.

Let me help you solve a few of these problems step-by-step:

General Form of Quadratic Equation

For any quadratic equation in the form: $ax^2 + bx + c = 0$

• The sum of the roots is given by: $\text{Sum} = -\frac{b}{a}$
• The product of the roots is given by: $\text{Product} = \frac{c}{a}$
• To find the roots, you can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Let's solve some of the equations from the image.

1. Equation 3: $x^2 - 8x - 9 = 0$

• Here, $a = 1$, $b = -8$, and $c = -9$.

Sum: $-\frac{b}{a} = -\left(-\frac{-8}{1}\right) = 8$

Product: $\frac{c}{a} = \frac{-9}{1} = -9$

Now, using the quadratic formula to find the roots: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-9)}}{2(1)} = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm \sqrt{100}}{2} = \frac{8 \pm 10}{2}$ So, the roots are: $x_1 = \frac{8 + 10}{2} = 9 \quad \text{and} \quad x_2 = \frac{8 - 10}{2} = -1$

2. Equation 4: $x^2 - 14x + 49 = 0$

• Here, $a = 1$, $b = -14$, and $c = 49$.

Sum: $-\frac{b}{a} = 14$

Product: $\frac{c}{a} = 49$

For the roots: $x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(49)}}{2(1)} = \frac{14 \pm \sqrt{196 - 196}}{2} = \frac{14 \pm 0}{2}$ Thus, the root is: $x_1 = x_2 = 7 \quad (\text{a double root})$

Let me know if you'd like more of these solved or a different problem from the sheet explained!

Here are 5 related questions:

1. What happens if the discriminant $(b^2 - 4ac)$ is negative?
2. How do we find the roots when the quadratic doesn't factor neatly?
3. Can the sum of the roots ever be zero? Under what conditions?
4. How do we determine the vertex of a parabola from its quadratic equation?
5. What are complex roots, and when do they occur?

Tip:

When solving quadratic equations, always check if the quadratic can be factored easily before using the quadratic formula. This can save time!