# A-level Mathematics/CIE/Pure Mathematics 1/Quadratics
## Completing the Square

Completing the square is a method of converting an expression of the form $ax^{2}+bx+c$ into an expression of the form $a(x-h)^{2}+k$.
It relies on the fact that $(x+p)^{2}=x^{2}+2px+p^{2}$.
First, we take the expression $ax^{2}+bx+c$ and factor out $a$ to get $a(x^{2}+(b/a)x+(c/a))$.
Then, we need to recognise that $(x+(b/2a))^{2}=x^{2}+(b/a)x+(b^{2}/4a^{2})$

### Examples

Suppose we need to convert $x^{2}+4x+3$ to completed-square form.
Here, $a$ is 1, so we don't need to do anything to factor it out.
Next, we recognise that $(x+2)^{2}=x^{2}+4x+4$, and need to find this in the expression.
$x^{2}+4x+3=x^{2}+4x+4-1=(x+2)^{2}-1$
Thus, we have our answer: $x^{2}+4x+3=(x+2)^{2}-1$

## The Discriminant

The discriminant is a value that we can use to determine how many real roots the quadratic function has. A real root is where the value of a quadratic expression is equal to zero.
The discriminant for an expression $ax^{2}+bx+c$ is calculated $\Delta =b^{2}-4ac$.
If the discriminant is greater than zero, there are two separate real roots.
If the discriminant is equal to zero, there is one reapeated root.
If the discriminant is less than zero, there are no real roots.

## Solving Quadratics

There are three main methods for solving a quadratic equation or inequality: factorising, completing the square, and using the quadratic formula.

### Factorising

Factorisation is where we break the expression into its factors.
e.g. $x^{2}+5x+6$ can be factorised as $(x+2)(x+3)$
Factorisation can be used to solve equations: if the product of two factors is equal to zero, that means that one of the factors has to be equal to zero.
e.g. Solve $x^{2}+x-6=0$
${\begin{aligned}x^{2}+x-6=(x+3)(x-2)=0\\\therefore x+3=0\quad or\quad x-2=0\\x=-3\ or\ 2\end{aligned}}$
To factorise an expression with a coefficient attached to the $x^{2}$ term, simply divide out the coefficient
e.g. Solve $3x^{2}+12x+9=0$
${\begin{aligned}3x^{2}+12x+9=0\\x^{2}+4x+3=0&\quad (\div 3)\\(x+3)(x+1)=0\\x=\{-3,-1\}&\quad \leftarrow {\text{Set notation can be used to represent multiple solutions}}\end{aligned}}$
However, not all expressions can be factorised.

### Completing the Square

Completing the square is where we convert a quadratic equation from the form $ax^{2}+bx+c$ to the form $a(x-h)^{2}+k$. This makes it easier to solve equations, and it works in all cases, unlike factorisation.
e.g. Solve $x^{2}+12x-9=0$
${\begin{aligned}x^{2}+12x-9=0\\(x+6)^{2}=x^{2}+12x+36\\(x+6)^{2}-36-9=0\\(x+6)^{2}=45\\x+6={\sqrt {45}}\\x=-6\pm 3{\sqrt {5}}\end{aligned}}$

### Quardatic Formula

The quadratic formula states that:
$ax^{2}+bx+c\iff x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\quad (a\neq 0)$
e.g. Solve $6x^{2}+x-9=0$
${\begin{aligned}6x^{2}+x-9=0\\x={\frac {-1\pm {\sqrt {1^{2}-4(6)(-9)}}}{2(6)}}\\x={\frac {-1\pm {\sqrt {1^{2}+216}}}{12}}\\x=(-1+{\sqrt {217}})/12\ or\ (-1-{\sqrt {217}})/12\end{aligned}}$
You may have noticed that the part under the square root is the discriminant. The reason this makes sense is that if the discriminant is negative, the square root cannot result in a real number, and thus there are no real roots. If the discriminant is zero, then $x=-b/2a$, so there is one repeated root. This leaves the case of when the discriminant is positive, resulting in two real roots.

## Solving Simultaneous Equations

Sometimes we will need to solve simultaneous equations which involve both a linear and a quadratic equation. To solve them, we need to use the method of substitution.
e.g. Solve the simultaneous equations $x+y=-1$ and $x^{2}+y^{2}=25$
${\begin{aligned}x+y=-1\rightarrow &\ y=x-1\\x^{2}+y^{2}=25\rightarrow &\ x^{2}+(x-1)^{2}=25\\x^{2}+x^{2}-2x+1=25\\2x^{2}-2x-24=0&\quad {\text{Now complete the square}}\\x^{2}-x-12=0\\(x-{\frac {1}{2}})^{2}=12+{\frac {1}{4}}\\x-{\frac {1}{2}}={\sqrt {\frac {49}{4}}}\\x={\frac {1}{2}}\pm {\frac {7}{2}}\\x=-3\ or\ 4\end{aligned}}$

## Recognising Quadratics

Sometimes quadratics will be hidden in other forms. If you can make a substitution to turn an expression into a quadratic, you can then solve it as you would a quadratic.
e.g. Find the value of x in $6x=1-{\sqrt {x}}$
${\begin{aligned}Let\ s={\sqrt {x}}\\6s^{2}=1-s\\6s^{2}+s-1=0\\s^{2}+{\frac {s}{6}}-{\frac {1}{6}}=0\\(s+{\frac {1}{12}})^{2}={\frac {1}{6}}+{\frac {1}{144}}\\s+{\frac {1}{12}}={\sqrt {{\frac {24}{144}}+{\frac {1}{144}}}}\\s={\frac {-1}{12}}\pm {\sqrt {\frac {25}{144}}}\\s={\frac {-1}{12}}\pm {\frac {5}{12}}\\s={\frac {-1}{2}}\ or\ {\frac {1}{3}}\\{\sqrt {x}}={\frac {-1}{2}}\ or\ {\frac {1}{3}}\\x={\frac {1}{4}}\ or\ {\frac {1}{9}}\end{aligned}}$
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