Solving quadratic word problems

How to solve word problems based on quadratic equations?

Solving word problems based on quadratic equations can be broken down into four simple steps:
  • Identity the unknown quantity and set it equal to a variable
  • Identify the algebraic conditions
  • Form a relevant quadratic equation from the above conditions
  • Solve the equation to get the value of the unknown quantity
For example, consider the following instance of a word problem based on quadratic equations:

An example

What is the area of a right triangle having a perimeter of 55 cm if the sum of its legs is 30 cm?
The solution to this word problem can be done in the three easy steps described above:
  1. Identity the unknown quantity and set it equal to a variable
    1. The unknown quantity is the area of the triangle. Let it equal to `A  cm^2` 
  2. Identify the algebraic conditions and form a relevant equation
    1. The algebraic conditions are that perimeter is equal to 55, and sum of the legs is 30 cm, and since it is a right triangle, the Pythagorean theorem as well. So if the three sides of the triangle are x, y and z, y being the hypotenuse, then we have the following equations corresponding to the given conditions:
      1. Perimeter is 55: `x + y + z = 55`
      2. Sum of legs is 30: `x + z = 30`
      3. Pythagorean theorem: `y^2 = x^2 + z^2`
  3. Form a relevant quadratic equation from the above conditions
    1. From the first two equations above we can clearly get the value of y by substituting the value of `x + z` in the first equation,
      1. `x + z + y = 55`
      2. `30 + y = 55`
      3. `y = 55 - 30`
      4. `y = 25`
    2. Substitute the value of `y` in the third equation,
      1. `y^2 = x^2 + z^2`
      2. `(25)^2 = x^2 + z^2`
    3. Solve the second equation either variable,
      1. `z = 30 - x`
    4. Substitute the above expression for x in equation 2(2) above,
      1. `(25)^2 = x^2 + (30 - x)^2`
  4. Solve the equation to get the value of the unknown quantity
    1. Change the equation to standard form
      1. `(25)^2 = x^2 + (30 - x)^2`
      2. `625 = x^2 + (30 - x)(30 - x)`
      3. `625 = 2x^2 - 60x + 900`
      4. `2x^2 - 60x + 275 = 0`
    2. Divide throughout by 2,
      1. `x^2 - 30x + 275 = 0`
    3. Solve by the quadratic formula,
      1. `x = (-b ± sqrt(b^2 - 4ac))/(2a)`
      2. `x = (30 ± sqrt((-30)^2 - 4*1*275))/(2*1)`
      3. `x = (30 ± sqrt(200))/2`
      4. `x = (30 ± 14.14)/2`
      5. `x = {(30 + 14.14)/2, (30 - 14.14)/2}`
      6. `x = {22.07, 7.93}`
Out of the above two solutions for `x`, the only positive one is taken to be the solution to for the word problem, since the variable `x` represents the length of the leg of a right triangle which can't be negative. So the length of one side is x = 22.07 cm. We already know the length of the hypotenuse y = 70 cm (see step 1 above) and from step 3, we know `z = 30 - x`, so `z = 30 - 22.07 = 7.93`
Therefore the dimensions of the triangle are 25, 22.07, 7.93.

No comments:

Post a Comment