         The Dragon's Demesne Examples - The Quadratic Formula             Home | Education | Recreation     Basic Quadratic Formula Examples               Example 1: Solve x^2 + 7x + 10 = 0 Solution: First we have to figure out what A, B and C are. We compare the equation given with the general form of the quadratic equation at the top of the page, Ax^2 + Bx + C = 0. B is 7 and C is 10, but what is A? In this case, A is 1. We do not usually write 1x^2, or 1x, we write x^2 or x instead, so if you see that, you know that there is a kind of 'invisible 1' just to the left of it. We put these values into the quadratic formula and see that: x ={ -(7) +/- sqrt[(7)^2 - 4(1)(10)]} / (2*1) If you do the math, you will then see that x = ( -7 +/- 3 ) /2 There are two possibilities for x, one is where we use the + in +/- and the other is where we use the - + case: x = ( -7 + 3 ) / 2 x = -2 - case: x = ( - 7 - 3 ) / 2 x = -5 The answers to the equation are x = -2 and x = -5. Try putting these numbers into the original equation as a check to make sure they work. Also, these are the only solutions to the equation; no other numbers will work. Example 2: Solve 2x^2 + 10x + 12 = 0 Solution: It should be easy to see by comparing this to Ax^2 + Bx + C = 0 that A=2, B=10, and C=12. Putting these into the quadratic formula gives: x = { -(10) +/- sqrt[(10)^2 - 4(2)(12)]} / (2*2) Working out these numbers, you will get that: x = ( - 10 +/- 2 ) / 4 + case: x = ( -10 + 2 ) /4 x = -2 - case: x = ( -10 - 2 ) / 4 x = -3 The answers are therefore x = -2 and x = -3. Example 3: Solve x^2 - 12x + 36 = 0. Solution: A=1, B=12, C=36. We put these into the quadratic formula. x = { -(-12) +/- sqrt[(-12)^2 - 4(1)(36)]} / (2*1) Working out the math, we get that: x = ( 12 +/- 0 ) / 2 Notice the 0 in this example. In this particular case, the + case and the - case will give the same value for x, which is that: + case and - case: x = 6 The reason this happened was because the term B^2 - 4AC equalled zero. Whenever this term is zero, the equation will only have one solution. Example 4: Solve 6x^2 - 294x = 0. Solution: This one might look scary, but it isn't. Compare it to the formula, and you will see that A=6, B=294, and C=0. (If you cannot see why C=0, it may help to notice that the equation can be though of as 6x^2 - 294x + 0 = 0) Put these into the quadratic formula, and we get that: x = { -(-294) +/- sqrt[(-294)^2 - 4(6)(0) ] } / (2*6) Evaluating this expression gives: x = (294 +/- 294) / 12 + case: x= (294 + 294)/12 x=49 - case: x = (294 - 294)/12 x=0 The solutions to this equation are x=49 and x=0. Example 5: Solve x^2 + 8x + 20 = 0 Solution: Put A=1, B=8, C=20 into the quadratic formula, and we get that: x = { -(8) +/- sqrt [(8)^2 - 4(1)(20)] } / (2*1) + case: x = [ -8 + sqrt (-16) ] / 2 - case: x = [ -8 - sqrt (-16) ] / 2 You may be wondering about the sqrt(-16) that showed up here. If you typed it into your calculator, you probably got an error that said something like "nonreal answer". When the quadratic formula results in the square root of a negative number in the solution, there is no answer to the question, and the solution is: x has no real roots. (It IS possible to find answers to this type of equation, but they require a higher level of math than was assumed. If you are interested in how to solve such equations, called complex equations, see the section on complex numbers.)] Questions (answers appear below) 1) 2) 3) 4) 5) 6) 7)               Enter supporting content here Please feel free to use any information you find on this site in any way you see fit.  Nothing here is copyrighted unless specifically noted.     