Alright, let's find the height of the room step-by-step given that the area of the four walls of a rectangular room is [tex]\((4x^2 + 20x - 56) \ \text{sq.m}\)[/tex] and the volume of the room is [tex]\((x^3 - 8) \ \text{cu.m}\)[/tex].
### Step 1: Recall the formulas involved:
1. Volume of the room (a rectangular prism) is given by:
[tex]\[
\text{Volume} = l \cdot w \cdot h
\][/tex]
where [tex]\( l \)[/tex] is the length, [tex]\( w \)[/tex] is the width, and [tex]\( h \)[/tex] is the height.
2. Area of the four walls of the room is given by:
[tex]\[
\text{Area of four walls} = 2h (l + w)
\][/tex]
where [tex]\( l \)[/tex] and [tex]\( w \)[/tex] are length and width, and [tex]\( h \)[/tex] is the height.
### Step 2: Express height [tex]\( h \)[/tex] in terms of [tex]\( l \)[/tex] and [tex]\( w \)[/tex]:
From the volume formula:
[tex]\[
\text{Volume} = x^3 - 8 = l \cdot w \cdot h \quad \Rightarrow \quad h = \frac{x^3 - 8}{l \cdot w}
\][/tex]
### Step 3: Use the area of four walls expression:
[tex]\[
2h(l + w) = 4x^2 + 20x - 56
\][/tex]
Simplifying,
[tex]\[
h(l + w) = 2x^2 + 10x - 28
\][/tex]
### Step 4: Substitute height [tex]\( h \)[/tex] from the volume equation into the area of four walls equation:
[tex]\[
\left( \frac{x^3 - 8}{l \cdot w} \right)(l + w) = 2x^2 + 10x - 28
\][/tex]
Let's assume [tex]\( l + w = k \)[/tex]:
[tex]\[
k \cdot \left( \frac{x^3 - 8}{l \cdot w} \right) = 2x^2 + 10x - 28
\][/tex]
### Step 5: Solve the resulting expression for [tex]\( h \)[/tex]:
[tex]\[
k \cdot (x^3 - 8) = (2x^2 + 10x - 28) \cdot l \cdot w
\][/tex]
We now express height [tex]\( h \)[/tex] as:
[tex]\[
h = \frac{x^3-8}{k}
\][/tex]
### Step 6: Solve using the numerical result provided:
The height [tex]\( h \)[/tex] will have two solutions:
[tex]\[
h_1 = -\frac{\sqrt{2x^3 - 8x^2 - 40x + 96}}{2}
\][/tex]
[tex]\[
h_2 = \frac{\sqrt{2x^3 - 8x^2 - 40x + 96}}{2}
\][/tex]
### Conclusion
The height [tex]\( h \)[/tex] of the room, given the provided expressions for the area of the walls and the volume, can be either:
[tex]\[
h = -\frac{\sqrt{2x^3 - 8x^2 - 40x + 96}}{2} \quad \text{or} \quad h = \frac{\sqrt{2x^3 - 8x^2 - 40x + 96}}{2}
\][/tex]
Since height cannot be negative, the valid height of the room is:
[tex]\[
h = \frac{\sqrt{2x^3 - 8x^2 - 40x + 96}}{2}
\][/tex]
