quadratic eqasions
...(x+4)=0, x is found easily. As (x+3)(x+4)=0, (x+3) or (x+4) MUST be equal to 0. Therefore: x+3 = 0 or x+4=0 so.. x=-3 or x=-4 Once we have found these answers, we have solved the quadratic. If constructing this quadratic on a graph, the values -3 and -4 would be the x-intercepts of the parabola. The parabola is what the quadratic graph is called. Another method of solving quadratics is called “completing the square”. When we expand (x + a) 2, we notice that it expands to x2 + 2ax + a2. So (x + 3) 2 expands to x + 6x + 9. In this equation we notice that the square root of the last number is half that of the number before the x. So… the square root of 9 is 3 and 3 is half of 6. In the example (x + 6) 2, the brackets expand to x2 + 12x + 36. The square root of 36 is 6 and 6 is half of 12. This tells us that this method works for all perfect squares. So when factorizing, the reverse is true. So with x2 + 12x + 36 we notice that half of 12 is 6 and the square root of 36 is 6. Therefore this equation must be a perfect square with (x + 6) 2. The perfect square form (x + a) + b, tells us that the vertex of the parabola is (-a,b). Example of completing the square: x2 + 6x + 12 As the square root of the last number is not half that of the number before the x, the equation must be changed. x2 + 6x + 9 -9+ 12 Because -9 and +9 equal 0, the equation has not changed. Now we can notice a perfect square…as the square root of 9… is half of 6. AO the equation factorizes to: (x + 3) 2 + 3 This is completing the square. From this form we can tell that the vertex of this parabola is at (-3,3). The two methods of factorizing shown above are very useful in different ways. However, not all quadratics can be factorized. One of the first things that we notice when looking at a factorized equation is that there are 2 parts: (x +...