A rectangular room has a width of $r$ m - Edexcel - A-Level Maths Pure - Question 10 - 2013 - Paper 2

The area $A$ of the room can be calculated using the formula:

$A = \text{length} \times \text{width}$

In this case, the area becomes:

$A = (r + 4) \times r = r^2 + 4r$

Given that the area is less than 21 m², we write the inequality as:

$r^2 + 4r < 21$

To solve the inequality $r^2 + 4r < 21$, we first rearrange it:

$r^2 + 4r - 21 < 0$

Next, we can factor the left-hand side:

$(r - 3)(r + 7) < 0$

Now, we find the critical points by setting each factor to zero:

1. $r - 3 = 0$ gives $r = 3$
2. $r + 7 = 0$ gives $r = -7$

To determine the intervals of $r$ that satisfy the inequality, we test values in the intervals $(-\infty, -7)$, $(-7, 3)$, and $(3, \infty)$:

• For $r < -7$: Choose $r = -8$: $(-)(-) > 0$ (not valid)
• For $-7 < r < 3$: Choose $r = 0$: $(+)(-) < 0$ (valid)
• For $r > 3$: Choose $r = 4$: $(+)(+) > 0$ (not valid)

Thus, the solution is:

$-7 < r < 3$

From part (a), we found that $r > 2.8$. From part (b)(ii), we found the inequality $-7 < r < 3$.

Combining these two results, we have:

1. $r > 2.8$
2. $r < 3$

Thus, the range of possible values for $r$ is:

$2.8 < r < 3$