EXAMPLE PROBLEMS QUADRATICS EQUATION

Example - Problem 1: A right triangle has a perimeter of 24 cm and a hypotenuse of 10 cm. Find the sides x and y, x > y, that make the right angle of the triangle.
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 D = b² - 4*a*c = 196 - 192 = 4
  
  
• Use the quadratic formulas to solve the quadratic equation; two solutions
  x1 = [ -b + sqrt(D) ] / 2*a = [ 14 + 2 ] / 2 = 8
  
  x2 = [ -b - sqrt(D) ] / 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 = sqrt (x² + y²)
  = sqrt (8² cm² + 6² cm²)
  = sqrt(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: 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.

Detailed Solution.

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Example - Problem 2: The sum of the squares of two consecutive real numbers is 61. Find the numbers.
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 D = b² - 4*a*c = 1 + 120 = 121

Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = [ -b + sqrt(D) ] / 2*a = [ -1 + 11 ] / 2 = 5
x2 = [ -b - sqrt(D) ] / 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.