> I have great problem with Quadratic Equations. Please > can any one of you tell me the rules in Quadratic > equations and how to solve it!
A quadratic equation is an equation is a second degree equation whose general form is Ax^2=Bx+C=0. For example, 2x^2+8x+6=0. Note that there can be equations not in this general form, but they can be put into this general form. For example, 2x^2+8x+9=3 can be put into the general form by simply subtracting 3 from both sides. Once the equation is in general form, it can be solved by one of 3 methods. The first is factoring. Notice that the above equation can be factored as (2x+2)(X+3)=0. Since one of these factors must equal 0 for the equation to equal 0, you can then set each factor equal to 0 and solve both of them. (Quadratic equations will always have 2 answers.) If we set 2x+2=0, we see that x can be -1. If we set X+3=0, we see x=-3. Plug -1 or -3 into the original equation, and you will see that they both work. The answer would be (-1,-3). Another way would be by completing the square. The first step in doing this is to move C to the right of the equation (in this case, 6.) Subtracting 6 from both sides gives us 2x^2+8x=-6. The next step is to divide both sides by A, in this case, 2. That gives us x^2+4x=-3. Then take half of the coefficient of the x term (here it is 4) , and square it. (4/2)^2=4. Add this square to both sides of the equation. Doing this gives us x^2+4x+4=1. The reason for doing all of these steps is to make the left side of the equation a perfect square. The equation can now be written (x+2)^2=1. It can be hard to see exactly why this works, and if you don't understand how creating perfect squares works, it is worth investigating further online as it is an application that is not only useful here, but in later mathematics. So in our problem, since we now have a perfect square, we can take the square root of both sides, giving us x+2=+/-sqrt 1. The right side is plus OR minus the square root of 1. Remember that when taking a square root, it will be both the positive and negative numbers of that root since a negative squared is positive also. For example, taking the square root of 4 gives us +2 and -2 since either one squared is 4. This is why quadratics have 2 answers. Back to our problem, x+2=+-sqrt 1. Subtracting 2 from both sides solves for x, and gives us x=-2+/-sqrt 1. The square root of 1 is just 1, so x=-2-1 or -2+1. x=(-1,-3). Quadratics can not always be solved by factoring, as they do not always factor. They can always be solved by completing the square, but problems are not often this neat and it can end up requiring far too much work. They can always be solved by the Quadratic Formula. This is done by taking A,B,and C from the original general equation, and placing it into the formula ((-B+/-sqrt(B^2-4AC))/2A. (read -B plus or minus the square root of(B squared -4AC) all divided by 2A. This equation is what you get when you solve Ax^2+Bx+C=0 for x. It is a hard problem that fortunately was solved hundreds of years ago, so now we just have to memorize the equation. The proof is quite interesting though, and is also a good exercise in completing the square so it is worth looking into.So for our equation, 2x^2+8x+6=0, we see that A=2, B=8, and C=6. Plugging this in we get (-8+/-sqrt(8^2-4(2)(6))/2(2) cleaning this up a bit, we get (-8+/-sqrt16)/4 or (-8+/-4)/4. so our 2 answers are (-8+4)/4 and (-8-4)/4, or (-4/4) and (-12/4) which is (-1,-3). Note that you will sometimes get answers that have the square root of a negative number, which are not real numbers, however they are real answers. they are imaginary numbers signified by i where i= sqrt(-1). That is kind of a general overview of quadratics. I hope it was helpful. If you have a specific question or problem, feel free to post it or e-mail me.