Considering there are multiple ways of solving a quadratic equation, there might be a method that you prefer over the others. You also need to take into consideration that some methods are not the best route to take depending on the equation. For example, if you have (x+5)2 you cannot use the quadratic formula with this equation without squaring the binomial. In my opinion, and through the Intellipath of my last math course I found the quadratic formula, factoring a quadratic equation, and factoring by grouping the simplest. Whereas, completing the square of a quadratic equation was very difficult for me to get through. I do not mind using the quadratic formula, because it is as simple as plugging in your values and solving the equation. So if you have y= x2+5x+4, we can use our quadratic equation format of y=ax2+bx+c, where a, b, and c are the coefficients and plug in our values to the formula to solve for x. Here is the quadratic formula, x=-b±b2-4ac2a . So now we just need to set our a, b, and c values. A=1 (Since there is no coefficient on the x2 it is a one), B=5, and C=4. So now we just plug in the values and solve.x=-(5)±(5)2-4(1)(4)2(1) → x=-(5)±25-162 → x=-5±92 → x=-5±32 →x=-5+32 =-1 x=-5-32 =-4 so x equals (– 4) or (– 1). Now if we were going solve this same equation by factoring, we need to ask ourselves what factors of 4 are the product of 5. Since 4 factors in to 4 and 1 and 2 and 2 this problem is fairly simple. We know that 2+2 doesn’t equal 5, but 4+1 does. So our binomials will be (x+1) and (x+4). We can double check this by multiplying the binomials to see if we come back to our equation. X multiplied by x is x2, x multiplied by 4 is 4x, x multiplied by 1 is x, and 1 multiplied by 4 is 4. So if we add that all together we get x2+5x+4. So now that we have our two binomials we can set them equal to 0 and solve for x. x+1=0, subtract 1 from both sides and you get x=-1. X+4=0, do the same and subtract 4 from both sides and your solution is x=-4. So we still come up with the same answers, x=-4 or x=-1.I find this the easiest method, even over the quadratic formula, because once you have figured out your binomials solving for x is a trivial computation. Now what if you have a quadratic equation where a≠1? It is not as simple to factor out a quadratic equation if the coefficient of x2 is not 1. This is where factoring by grouping comes into play. There is a rule that will help you solve a quadratic equation where A≠1, it implies that if A≠1, A multiplied by C is equal to two numbers added together. So let us say our equation is 3x2+7x+4. So let us set our coefficients to A, B, and C. A=3, B=7, C=4 So if A multiplied by C is 12 we need to find factors of 12 are the sum of 7, just like we did above. 12 factors into 12 and 1, 6 and 2, and 4 and 3. Obviously the only factors that will add to 7 are 4 and 3. An easy way to look at this is that the 7x is just 3x + 4x. Now rewrite your equation as 3x2+3x+4x+4. Now we need to group the equation as (3x2+3x) + (4x+4). Now we can factor these out individually by pulling 3x out of (3x2+3x) to get 3x(x+1), and pulling 4 out of (4x+4) to get 4(x+1). Now we are left with 3x(x+1) + 4(x+1) which we can now simplify to (3x+4)(x+1). I think this is the most difficult out of the three methods so far, because it entails a significant amount of more work. Now for the method of completing the square, let’s use the equation x2 + 10x = 24. Before we start there is a rule that B divided by two will equal C2. So in this equation we have our A and B but our c is unknown at this point. I find this method the hardest of the four because I always wanted to just subtract 24 from both sides and factor it from there. In this equation that is possible but sometimes that is not the case. So if we divide 10 by 2 we get 5. So if ½ of B is C2 we can add our C to our equation now, but we must add it to both sides of the equation. This will give us x2 +10x + (5)2=24 + (5)2 →x2 + 10x + 25 = 49. Now we can factor out the left side of our equation. Ask the question what plus what equals B(10) and multiplies to C(25)? This will give us our binomials of (x+5)(x+5) = 49, which we can write as (x+5)2 = 49. Now we can take the square root of both sides of the equation (x+5)2=49. Since we took the square root of both sides we need to add a ± to the opposite side of the variable. So this simplified a little further will be x+5=±7. Now we solve just like we do with the quadratic equation to find our two answers. Subtract 5 from both sides and we have two different solutions x= -5+7→x=2 or x=-5-7→x=-12. So our answers are x=2, or -12. A real world example of a quadratic equation could be finding the length and width of a building if all you know is the area, and that the width is 2 feet shorter than the length. Let’s say the area is 360 ft2, if length multiplied by the width is the area, we can make a quadratic equation to find out what the length and the area is. To begin since we know that the width is 2 ft. shorter than the length we will let (x+2) represent the length, and (x) represent the width. So now we have x(x+2) = 360. Let’s subtract 360 from both sides and multiply x by (x+2). Now we have a quadratic equation of x2 + 2x – 360 = 0. If we factor it we get (x+20)(x-18) = 0 so our answers will be x = 18, and x = -20 but since we know that we cannot have a negative length it is 20 ft., and the width is 18ft. This quadratic equation lets you find the dimensions of a rectangle if you are given the difference between the length and width as well as the area. To answer how you can use the solutions from the quadratic equation of x2 - 8x + 12, which are x=2 and x=6, we can refer back to the basic format of a quadratic equation which is Ax2+Bx+C=y. So when we set y to 0, we are solving for the x-coordinate that is in correspondence to the y-coordinate of 0. Thus, if we know that these x-coordinates have a y-coordinate of 0, then they are in fact the x-intercept. So for this equation this is a parabola that has two coordinates of (2,0), and (6,0) where they cross the x-axis.