Detailed Solutions to Matched Problems

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
    2x + 2y = 60

  • The area of the rectangle is 200 m2, hence
    x*y = 200

  • Solve the equation 2x + 2y = 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 +30x - 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 + sqrt(D) ] / 2*a = [ -30 + 10 ] / -2 = 10 m

    x2 = [ -b - sqrt(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 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.
    x2 + 2x - 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 + sqrt(D) ] / 2*a = [ -2 + 10 ] / 2 = 4
    x2 = [ -b - sqrt(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 on how to Solve Equations, Systems of Equations and Inequalities.