Quadratic Equations Long Questions (Question 14 & 15)

Question 14 (4 marks):
An aquarium has the length of (x + 7) cm, the width of x cm and the height of 60 cm.
The total volume of the aquarium is 48000 cm³. The aquarium will be filled fully with water.
Calculate the value of x.

Solution:


$\begin{array}{l}\text{Volume of aquarium}=48000{\text{ cm}}^3\\ \text{So,}\\ \left(x\right)\left(x+7\right)\left(60\right){\text{ cm}}^3=48000{\text{ cm}}^3\\ \left(x\right)\left(x+7\right)=\frac{48000}{60}\\ x^2+7x=800\\ x^2+7x-800=0\\ \left(x-25\right)\left(x+32\right)=0\\ x-25=0\text{ or }x+32=0\\ x=25\text{ or }x=-32\left(\text{not accepted}\right)\\ \\ \text{So the value of }x=25\text{ cm}\end{array}$

Question 15 (4 marks):
Diagram 3 shows a garden path with a rectangular shape. There are 8 similar circular stepping stone built in the path.


Given the area of the path is 32 m², find the diameter, in m, of one piece of stepping stone.

Solution:
$\begin{array}{l}\text{Given the area of the path}\\ =32{\text{ m}}^2=320000{\text{ cm}}^2\\ x\left(x+4\right)=320000\\ x^2+4x-320000=0\\ a=1,b=4,c=-320000\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-4\pm \sqrt{4^2-4\left(1\right)\left(-320000\right)}}{2\left(1\right)}\\ x=\frac{-4\pm \sqrt{1280016}}{2}\\ x=\frac{-4+1131.38}{2}\text{ or }\frac{-4-1131.38}{2}\\ x=563.69\text{ or }-567.69\left(\text{ignored}\right)\\ \\ \text{Diameter of one piece of stepping}\\ \text{stone}=x\\ =563.69\text{ cm}\\ =5.64\text{ m}\end{array}$