![]() ![]() The techniques of solution were further refined by the Greeks, the Arabs and Indians, and finally a complete and coherent treatment was completed once the notion of complex numbers was understood. The history of quadratics will be further explored in the History section, but we note here that these types of equations were solved by both the Babylonians and Egyptians at a very early stage of world history. Surprisingly, when mathematics is employed to solve complicated and important real world problems, quadratic equations very often make an appearance as part of the overall solution. Both in senior mathematics and in tertiary and engineering mathematics, students will need to be able to solve quadratic equations with confidence and speed. While quadratic equations do not arise so obviously in everyday life, they are equally important and will frequently turn up in many areas of mathematics when more sophisticated problems are encountered. In this module we will develop a number of methods of dealing with these important types of equations. The rearrangements we used for linear equations are helpful but they are not sufficient to solve a quadratic equation. We keep rearranging the equation so that all the terms involving the unknown are on one side of the equation and all the other terms to the other side. The essential idea for solving a linear equation is to isolate the unknown. The equation = is also a quadratic equation. Thus, for example, 2 x 2 − 3 = 9, x 2 − 5 x + 6 = 0, and − 4 x = 2 x − 1 are all examples of quadratic equations. Roughly speaking, quadratic equations involve the square of the unknown. ![]() Such equations arise very naturally when solving elementary everyday problems.Ī linear equation involves the unknown quantity occurring to the first power, thus, This is quite difficult to type out but easy to actually use.In the module, Linear Equations we saw how to solve various types of linear equations. To solve this using the quadratic formula, the integers just have to be subbed into the following equation: In the following example a, b, and c represent the integers in front of each part of the quadratic. The last way of solving a quadratic is using the quadratic formula. ![]() To solve for x we just add 5 to both sides and take the square root. To fix this we just take off another 5 after our squared bracket giving us a final equation of If we expand the squared bracket we get the x 2 and the 10x that we need, but we get a +25, when we need +20. This is easier shown than explained with words.įirst, the coefficient of x (the ten infront of the x) is halved, and this is the constant used in the bracket with x.īut we want (x+5) 2 + a constant to be equal to x 2+10x+30 = 0. To 'complete the square' of a quadratic, the initial equation is rewritten as a (x + constant) bracket squared minus another constant to give the same value as the starting equation. (ax+b)(cx+d) = 0, where a,b,c and d are integers.Ī * c must equal 1 to give us the original 1x 2.Ī * d + b * c must be equal to 5 to give us 5x.Īnd b * d must be equal to 6 to give us our constant. So in order to split the equation into two brackets we have to know which numbers are needed. ^^^ when these brackets are multiplied out they give the original equation. If a common factor cannot be found, the next step is to try and put the equation into two brackets that are multiplied together. Rearranging these equations gives us the final solutions of This would then be solved by setting each part equal to zero, In the case a factor of 2x can be taken out, making the equation look like this: The first step for factorisation is to see if a common factor can be taken out, this is the easiest way of solving a quadratic. For your maths GCSE it is important that you understand the three main methods of solving quadtratics: factorisation, completing the square, and using the quadratic formula. ![]()
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