Using quadratic equations to solve problems; detailed solutions and explanations are included.

Problems with Solutions

Problem 1:

Solution to Problem 1:

• We start by drawing a triangle with the given information
  
• The perimeter of the triangle is 24, hence
  x + y + 10 = 24
  
• It is a right triangle, use Pythagoras theorem to obtain.
  x² + y² = 10²
  
• Solve the equation x + y + 10 = 24 for y.
  y = 14 - x
  
• Substitute y in the equation x² + y² = 10² by the expression obtained above.
  x² + (14 - x)² = 10²
  
• Expand the square, group like terms and write the above equation with the right side equal to zero.
  2x² -28x + 96 = 0
  
• Multiply all terms in the above equation by 1/2.
  x² -14x + 48 = 0
  
• Find the discriminant of the above quadratic equation.
  Discriminant Δ = b² - 4*a*c = 196 - 192 = 4
  
• Use the quadratic formulas to solve the quadratic equation; two solutions
  x1 = [ -b + √Δ ] / (2 a) = [ 14 + 2 ] / 2 = 8
  x2 = [ -b - √Δ ] / (2 a) = [ 14 - 2 ] / 2 = 6
  
• use the equation y = 14 - x to find the corresponding value of y.
  y1 = 14 - 8 = 6
  y2 = 14 - 6 = 8
  
• Taking into account the condition x > y, the sides that make the right angle of the triangle are: x = 8 cm and y = 6 cm.
  Check answer:
  Hypotenuse h = √ (x² + y²)
  = √ (8² cm² + 6² cm²)
  = √(64 cm² + 36 cm²)
  = 10 cm, it agrees with the given value.
  Perimeter = y + x + hypotenuse
  = 8 cm + 6 cm + 10 cm
  = 24 cm, it agrees with the given value.

Matched Problem 1:

Solutions to Matched Problems

Problem 2:

Solution to Problem 2:

• Let x and x+1 be the two consecutive numbers. The sum of the square of x and x + 1 is equal to 61.
  x² + (x + 1)² = 61
  
• Expand (x + 1)², group like terms and write the above equation with the right side equal to zero.
  2x² + 2x - 60 = 0
  
• Multiply all terms in the above equation by 1/2.
  x² + x - 30 = 0
  
• Find the discriminant of the above quadratic equation.
  Discriminant Δ = b² - 4*a*c = 1 + 120 = 121
  
• Use the quadratic formulas to solve the quadratic equation; two solutions
  x1 = [ - b + √Δ ] / 2*a = [ -1 + 11 ] / 2 = 5
  x2 = [ - b - √Δ ] / 2*a = [ -1 - 11 ] / 2 = -6
  
• First solution to the problem
  first number: x1 = 5
  second number: x1 + 1 = 6
  
• Second solution to the problem
  first number: x2 = -6
  second number: x2 + 1 = -5
  Check answer:
  first solution sum of squares: 5² + 6²
  = 25 + 36 = 61
  second solution sum of squares: (-6)² + (-5)²
  = 36 + 25 = 61
  The two solutions to the problem agree with the given information in the problem.
  

Matched Problem 2:

Solutions to Matched Problems

Solutions to Matched Problems

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.

Solution to Matched 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.

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