I would probably say this is around the half way point in IGCSE Mathematics, so if you have made it this far well done! So we’ve briefly covered what quadratic equations are already and how to factorise and expand them as well as completing the square. We are now going to look at a cheat way to solve them which requires you to memorise another long equation!
A really really easy way to solve quadratic equations is through the quadratic formula which is a fairly long equation rearranged from the general form of a quadratic to make x by itself on one side and all the other terms on the other side.
$$ ax^2 + bx + c \Rightarrow x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
This makes solving quadratic equations really easy… however… if an equation asks you to factorise you must factorise… (unfortunately). Let’s try using the equation on a quadratic:
$$ 2x^2 - 3x + 1 = 0 \Rightarrow x = \dfrac{-(- 3) \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)} $$
$$ x = \dfrac{3 \pm \sqrt{1}}{4} \therefore x = \dfrac{4}{4}, \dfrac{1}{2} $$
See that we are able to get our 2 answers since we either add or subtract the square root bit.
We also need to be able to solve simultaneous equations with one linear equation and one quadratic equation. For this it’s probably best to use the substitution method although you could use the elimination method if you were to get the same letter term (to the same power) on both equations.
Let’s try this simultaneous equation:
$$ y = 2x - 11 \color{blue}{\text{ (1) }} $$ $$ x^2 + y^2 = 25 \color{blue}{\text{ (2) }} $$ $$ x^2 + y^2 \color{red}{-x^2} = 25 \color{red}{-x^2} $$ $$ y^2 = 25 - x^2 \color{blue}{\text{ (2a) }} $$ $$ (y)\color{red}{^2} = (2x - 11)\color{red}{^2} \color{blue}{\text{ (1a) }} $$ $$ \color{blue}{\text{ (1a into 2a) }} $$ $$ 25 - x^2 = (2x - 11)^2 = (2x - 11)(2x - 11) = 2x \color{red}{\times 2x} - 11 \color{red}{\times 2x} + 2x \color{red}{\times -11} - 11 \color{red}{\times -11} $$ $$ 25 - x^2 \color{red}{-25 + x^2} = 4x^2 - 22x - 22x + 121 \color{red}{-25 + x^2} $$ $$ 0 = 5x^2 - 44x + 96 $$
Now that we have a quadratic equation with one side being equal to 0, we can solve using any of the methods we have learnt. For this, I will solve by factorisation (putting the equation into 2 brackets):
$$ 0 = (x - 4)(5x - 24) \therefore x_1 = 4, x_2 = \dfrac{24}{5} = 4.8 $$
Now we can solve by substituting the value for x into y. We have 2 values for x so we need to do this process twice (it’s recommended to use the linear equation when possible for substitution): $$ y_1 = 2\color{red}{(4)} - 11 \therefore y_1 = -3 \ \color{blue}{ ( x_1 \text{ into 1}) } $$ $$ y_1 = \color{red}{\tfrac{48}{5}} - \color{red}{\tfrac{55}{5}} \therefore y_1 = \dfrac{-7}{5} \ \color{blue}{ ( x_2 \text{ into 1}) } $$
Graphically, these lines must intersect at 2 different points, why? Because the 2nd equation is actually the equation for a circle (we don’t need to know this at GCSE but it is quite cool):
More coming soon