Solve problems involving use of quadratic equations
Example 1.
![Picture](/uploads/1/1/5/3/11531538/9056317.png)
Question: The
length of a field is 2 metres more that 3 times the breadth. Its area is 385
square metres. Find its perimeter
Solution: let the length be x and breadth be y. Therefore x = (3y + 2) metres. We also know that xy = 385 square metres. Substitute to get (3y + 2)y = 385 square metres. Expanding this equation will give us a quadratic equation as shown on the left. Factorising this will give us (x - 11)(3x + 35) = 0. Therefore x = 11 or x = -35/3. [Note: as x is a length, it must be positive. It is important to write x>0 if it's a length.] Therefore x = 11. y = 385/11 = 35. Therefore the length is 11 metres and the breadth is 35 metres. Therefore the perimeter is (2 x 11) + (2 x 35) = 22 + 70 = 92 metres.
Solution: let the length be x and breadth be y. Therefore x = (3y + 2) metres. We also know that xy = 385 square metres. Substitute to get (3y + 2)y = 385 square metres. Expanding this equation will give us a quadratic equation as shown on the left. Factorising this will give us (x - 11)(3x + 35) = 0. Therefore x = 11 or x = -35/3. [Note: as x is a length, it must be positive. It is important to write x>0 if it's a length.] Therefore x = 11. y = 385/11 = 35. Therefore the length is 11 metres and the breadth is 35 metres. Therefore the perimeter is (2 x 11) + (2 x 35) = 22 + 70 = 92 metres.