Quadratic Equations - Problems (1)

This is a tutorial on using the quadratic equations to solve problems. Detailed solutions and explanations are included.


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
    Triangle to solve
  • 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:

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.


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 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:

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

Detailed Solution.

More References and links

Solve Equations, Systems of Equations and Inequalities.