Quadratics

$ax^2 + bx + c = 0$

All quadratics have this general equation, however you can just remember them by having an $x^2$ term.

The coefficient of a number is the number in front of the $x$ (the amount you are multiplying it by). For example $a$ is the coefficient of $x^2$.

Solving by factorising

This is similar to factorising "linear" equations, but instead of putting the equation into 1 bracket we need 2.

For example lets solve $x^2 - x = 12$ Before we factorise we must make 0 equal to the equation on the other side:

$x^2 - x - 12 = 0$ There are several different methods of factorising quadratics.

"Guessing Route" (when $a = 1$)

$x * y = ac$ and $x + y = b$

because of the 1st rule we can shorten our list to factors of $c$ then we can see if those satisfy the second rule as well or not. Let's demonstrate this using our example:

$\begin{aligned} \text{factors of 12:}& \\ 12 * 1 &= 12 \\ 6 * 2 &= 12 \\ 4 * 3 &= 12 \\ 3 * 4 &= 12 \\ \end{aligned}$

Note that $ac =1 * -12$ but I am finding factors of 12.

From the equation, $b = -1$ since $-1 * x = -x$.

We can now see that $3 - 4 = -1$ so we have found our 2 numbers. Since $a = 1$ we can easily just add these 2 numbers to different brackets with $x$ in each:

$(x - 4) (x + 3) = 0$

We should now check that this is correct by multiplying out the brackets to see if it gives the answer the same as the original quadratic equation.

After that, the solutions for $x$ are the numbers that make one of the brackets equal to $0$. So:

$x = 0$

Factoring by grouping (when a is not 1)

We are using the same 2 rules that the other method uses, however we are trying to factorise one bracket at a time. For this example lets use this equation:

$3x^2 - 7x - 6 = 0$

The first step is the same, we find the 2 pairs of numbers that satisfy the 2 rules: $x * y = 3 * -6 = -18$ $x \pm y = -7$

Lets start by getting our factors of 18:

$\begin{aligned} 18 * 1 &= 18 \\ 9 * 2 &= 18 \\ 6 * 3 &= 18 \\ 3 * 6 &= 18 \end{aligned}$

From this we can see that $-9 + 2$ will make $-7$. We now break up the $b$ term into the 2 numbers:

$3x^2 \color{red}{-9x + 2x} - 6 = 0$

We now look for the highest common factor in the first pair of numbers: $\color{blue}{3x^2 - 9x}$.

The highest common factor of 3 and 9 will be 3. Also, both terms have $x$ so the highest common factor is $3x$:

$3x (x - 3)$

The bit in the brackets will the same for the other pair of terms, so we can factorise the whole thing easily now:

$3x(x-3) + 2(x - 3) = 0$

So now $(3x + 2)$ make one bracket and $(x - 3)$ makes another bracket; so we have fully factorised the quadratic:

$(3x + 2)(x - 3) = 0$

To answer the question we need to make each bracket equal to $0$ separately. So the first bracket would be:

$3x + 2 = 0 \rightarrow 3x = -2 \rightarrow x = \dfrac{-2}{3}$

The other answer would be 3.

We can now check and make sure it expands to make the original quadratic.

Quadratic Formula

Another way to solve a quadratic is through a complicated formula:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

All we have to do is put the numbers in the right place and type it into a calculator! The $\pm$ part provides us with our 2 answers. For example lets use the example above to show how to solve it this way:

$\begin{aligned} x &= \dfrac{-\color{red}{(-7)} \pm \sqrt{\color{red}{-7^2}-4(\color{red}{3 * -6})}}{2\color{red}{* 3}} \\[7mu] x &= \dfrac{7 \pm \sqrt{121 }}{6} = \dfrac{7 \pm 11}{6} \\ x &= \dfrac{7 + 11}{6} = 3 \text{ OR } \dfrac{7 - 11}{6} = \dfrac{-2}{3} \end{aligned}$