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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.