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 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.


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.

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 on how to Solve Equations, Systems of Equations and Inequalities.







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Updated: 2 April 2013

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