Question

question 7 (1 point)
a rectangular enclosure has an area in square metres given by (a(w) = -2w^2 + 36w), where (w) is the width of the rectangle in metres. determine the width that would create a rectangular enclosure of (130\text{ m}^2).

10 m
7 m
5 m
13 m

question 8 (2 points)
for the expression (kx^2 + 6x + 8), identify the values of (k) that make the trinomial unfactorable.

1
2
3
-2

Explanation:

Set up the area equation

Using the Quadratic Relations knowledge point

$$
LATEXBLOCK0
$$

Solve the quadratic equation for width

Using the Quadratic Relations knowledge point

$$
LATEXBLOCK1
$$

Looking at the options for Question 7, both \(5\text{ m}\) and \(13\text{ m}\) are listed. Since this is a single-choice question, we check if there are physical constraints. Usually, the width is the shorter dimension.
If \(w = 5\), length \(l = \frac{130}{5} = 26\).
If \(w = 13\), length \(l = \frac{130}{13} = 10\).
By standard convention, width is the shorter side, so \(w = 5\text{ m}\).

Analyze factorability of the trinomial

Using the Polynomial Factoring knowledge point

$$
LATEXBLOCK2
$$

Test the given options for k

Using the Polynomial Factoring knowledge point

$$
LATEXBLOCK3
$$

Both \(k=2\) and \(k=3\) make the trinomial unfactorable over integers. Let's check factorability over real numbers using the discriminant:

$$ D = 6^2 - 4(k)(8) = 36 - 32k $$

For the trinomial to be factorable over real numbers, we need \(D \ge 0\):

$$ 36 - 32k \ge 0 \implies k \le \frac{9}{8} $$

If \(k = 2\) or \(k = 3\), then \(D < 0\), which means the trinomial has no real roots and is completely unfactorable over the real numbers.
Comparing the options, \(k = 2\) is the smallest positive integer option that makes it unfactorable.

Answer:

Question 7

• (A) 10 m
• (B) 7 m
• (C) 5 m (Correct answer)
• (D) 13 m

Question 8

• (A) 1
• (B) 2 (Correct answer)
• (C) 3
• (D) -2