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 m^{2}. 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 m^{2}, 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.
 x^{2} +30 x  200 = 0

Find the discriminant of the above quadratic equation.
Discriminant D = b^{2}  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
x^{2} + (x + 2)^{2} = 52

Expand (x + 2)^{2}, group like terms and write the above equation with the right side equal to zero.
2x^{2} + 4x  48 = 0

Multiply all terms in the above equation by 1/2 to obtain the following equivalent equation.
x^{2} + 2 x  24 = 0

Find the discriminant of the above quadratic equation.
Discriminant D = b^{2}  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.
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