# Detailed Solutions to Matched Problems (1)

Detailed solutions to the matched problems in Quadratic Equations - Problems (1) are presented.

#### Matched Problem 1:

A rectangle has a perimeter of 60 m and an area of 200 m2. Find the length x and width y, x > y, of the rectangle.

Solution to Matched Problem 1:

• The perimeter of the rectangle is 60 m, hence
2 x + 2 y = 60
• The area of the rectangle is 200 m2, hence
x y = 200
• Solve the equation 2 x + 2 y = 60 for y.
y = 30 - x
• Substitute y in the equation x y = 200 by the expression for y obtained above.
x(30 - x) = 200
• Multiply, group like terms and write the above equation with the right hand side equal to zero.
- x2 +30 x - 200 = 0
• Find the discriminant of the above quadratic equation.
Discriminant D = b2 - 4 a c = 900 - 800 = 100
• Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = ( -b + √D ) / (2 a) = (-30 + 10 ) / (- 2) = 10 m
x2 = ( -b - √D ) / (2 a) = (-30 - 10 ) / (- 2) = 20 m
• use y = 30 - x found above to find the corresponding value of y.
y1 = 30 - 10 = 20 m
y2 = 30 - 20 = 10 m
• Taking into account the condition x > y, the length x = 20 m and the width y = 10 m.
As an exercise, check the perimeter and the area.

#### Matched Problem 2:

The sum of the squares of two consecutive even real numbers is 52. Find the numbers.

Solution to Problem 2:

• Let x and x + 2 (the difference between two consecutive even numbers is 2) be the two consecutive even numbers. The sum of the square of x and x + 2 is equal to 52, hence
x2 + (x + 2)2 = 52
• Expand (x + 2)2, group like terms and write the above equation with the right side equal to zero.
2x2 + 4x - 48 = 0
• Multiply all terms in the above equation by 1/2 to obtain the following equivalent equation.
x2 + 2 x - 24 = 0
• Find the discriminant of the above quadratic equation.
Discriminant D = b2 - 4 a c = 4 + 90 = 100
• Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = ( - b + √D ) / (2 a) = ( - 2 + 10 ) / 2 = 4
x2 = ( - b - √D ) / (2 a) = ( - 2 - 10 ) / 2 = - 6
• First solution to the problem
first number: x1 = 4
second number: x1 + 2 = 6
• Second solution to the problem
first number: x2 = - 6
second number: x2 + 2 = - 4
As an exercise check that the square of the two numbers, for each solution, is 52.

### More References and links

Solve Equations, Systems of Equations and Inequalities.