Question

③ vertex form: f(x) = (x - h)² + k h=
k= vertex =
⑪ x² + 18x +
⑫ x² + 5x - 14 = 0
⑨ x² + 6x - 9 = 0
⑦ f(x) = 3(x - 2)³ + 6
f(x)
⑩ two consecutive number have
a product of 600
the equation is
the two numbers are and

Explanation:

Step1: Identify the problem type

The problem involves quadratic equations and vertex form, which falls under Algebra (a subfield of Mathematics). Let's take problem ⑨: \(x^2 + 6x - 9 = 0\) and solve it by completing the square (to find vertex form or solve the equation).

Step2: Complete the square for \(x^2 + 6x - 9 = 0\)

First, move the constant to the other side: \(x^2 + 6x = 9\).
Take half of the coefficient of \(x\) (which is \(6\), half is \(3\)), square it (\(3^2 = 9\)), and add to both sides: \(x^2 + 6x + 9 = 9 + 9\).
This factors to \((x + 3)^2 = 18\).
To solve for \(x\), take square roots: \(x + 3 = \pm\sqrt{18} = \pm3\sqrt{2}\).
So \(x = -3 \pm 3\sqrt{2}\).

Step3: For vertex form (problem ③)

The vertex form of a quadratic is \(f(x) = a(x - h)^2 + k\), where \((h, k)\) is the vertex. For a quadratic \(ax^2 + bx + c\), \(h = -\frac{b}{2a}\) and \(k = f(h)\).

Step4: Problem ⑩ (two consecutive numbers with product 600)

Let the first number be \(x\), the next consecutive number is \(x + 1\). Their product is \(x(x + 1) = 600\), which simplifies to \(x^2 + x - 600 = 0\).
Factor (or use quadratic formula): \(x^2 + x - 600 = (x + 25)(x - 24) = 0\) (since \(25 \times 24 = 600\) and \(25 - 24 = 1\)). So \(x = 24\) or \(x = -25\). Thus the numbers are \(24\) and \(25\) (or \(-25\) and \(-24\)).

Answer:

For problem ⑨: \(x = -3 \pm 3\sqrt{2}\)
For problem ⑩: The equation is \(x(x + 1) = 600\) (or \(x^2 + x - 600 = 0\)), and the numbers are \(24\) and \(25\) (or \(-25\) and \(-24\))
(Note: Depending on the specific problem focused on, the answer can be adjusted. Here we covered a few visible problems.)