Let’s solve the provided problems step by step.

8. Consider the equation $x^2 - 6x + 9 = 0$.

Part (a): Find the discriminant

The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by: $\Delta = b^2 - 4ac$

Here, $a = 1$, $b = -6$, and $c = 9$. Substituting these values: $\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0$

So, the discriminant is 0.

Part (b): Number of roots

If:

• $\Delta > 0$: Two real and distinct roots.
• $\Delta = 0$: One real and repeated root.
• $\Delta < 0$: Two complex roots.

Since $\Delta = 0$, the equation has one real root.

Part (c): Axis of symmetry

The axis of symmetry for a quadratic equation $y = ax^2 + bx + c$ is given by: $x = -\frac{b}{2a}$

Substituting $a = 1$ and $b = -6$: $x = -\frac{-6}{2(1)} = \frac{6}{2} = 3$

So, the axis of symmetry is $x = 3$.

9. Solve the system of equations:

$y = x^2 \quad \text{and} \quad y = x + 6$

Substitution Method

Since both equations equal $y$, set them equal to each other: $x^2 = x + 6$

Rearrange to form a standard quadratic equation: $x^2 - x - 6 = 0$

Factorize the quadratic: $x^2 - x - 6 = (x - 3)(x + 2) = 0$

Solve for $x$: $x = 3 \quad \text{or} \quad x = -2$

Find the corresponding $y$-values:

• For $x = 3$: $y = 3^2 = 9$.
• For $x = -2$: $y = (-2)^2 = 4$.

Thus, the solutions are: $(3, 9) \quad \text{and} \quad (-2, 4)$

Final Answers:

1. 8a: Discriminant = 0
2. 8b: One real root
3. 8c: Axis of symmetry = $x = 3$
4. 9: Solutions are $(3, 9)$ and $(-2, 4)$

Would you like detailed steps or explanations on any part?

Here are five related questions:

1. How do you determine the nature of roots based on the discriminant?
2. Why does the axis of symmetry always pass through the vertex of a parabola?
3. How would the solutions change if $\Delta > 0$ in the quadratic equation?
4. Can you solve the same system of equations graphically? What would the intersection points look like?
5. How does substituting into the second equation verify the solutions?

Tip: Always double-check discriminant calculations to ensure the correct classification of roots!