Algebra

Laws of indices

Multiplication: $x^{a} \times x^{b} = x^{a + b}$
Division: $x^{a} \div x^{b} = x^{a - b}$
Exponentiation: $(x^{a})^{b} = x^{ab}$
Fractional exponents: $x^{1/ a} = \root a \of x$
Negative exponents $x^{- a} = \frac{1}{x ^{a}}$

Rationalising the denominator

To rationalise the denominator of a fraction, multiply by the conjugate radical of the denominator (e.g. $a - b c$ for $a + b c$ ) to create a difference of two squares.

Quadratics

A quadratic equation is a polynomial with degree 2. The general form is:

a x^{2} + b x + c = 0

Quadratics can be solved by factorising. A quadratic with roots $r$ and $s$ can be factorised as $a (x - r) (x - s)$ . The roots, of the quadratic are where the expression equals zero.

Quadratic formula

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The quadratic formula can be derived by completing the square on a general quadratic $a x^{2} + b x + c = 0$ . A proof is provided below.

Discriminant

The discriminant of a quadratic equation with general equation $a x^{2} + b x + c = 0$ is $b^{2} - 4 a c$ . If the discriminant is:

negative, then there are no real roots.
0, then there is one repeated real root.
positive, then there are two distinct real roots.

If the discriminant is 0, then there is one repeated real root, which is:

x = \frac{- b \pm b ^{2} - 4 a c}{2 a} = \frac{- b \pm 0}{2 a} = \frac{- b}{2 a}

If the quadratic has rational coefficients and $x + y z$ is a root, then $x - y z$ (the conjugate radical) is a root.

The line of symmetry is at $x = \frac{- b}{2 a} = \frac{x _{1} + x _{2}}{2}$ . Quadratics can be put into completed square form to find their line of symmetry and vertex:

Hidden quadratics

Quadratics can also be hidden by being quadratic in a function. For example, the below are all hidden quadratics:

3 e^{2 x} - 11 e^{x} + 6 cot^{2} x - 5 cot x x^{4} + 4 x^{2} + 4 = 0 = 6 = 0

These hidden quadratics can all be solved by doing a substitution (e.g. let $y = e^{x}$ ), or just directly factorising as if $f (x)$ was just $x$ .

Polynomials

IS: expanding brackets, collecting like terms, factorising, simple algebraic division

Factor theorem

The factor theorem gives a connection between the roots of a polynomial and its factorisation. The two forms are:

f (a) = 0 f (\frac{b}{a}) = 0 ⟺ (x - a) is a factor of f (x) ⟺ (a x - b) is a factor of f (x)

Proofs

Quadratic formula

The quadratic formula can be derived as shown:

a x^{2} + b x + c x^{2} + \frac{b x}{a} + \frac{c}{a} (x + \frac{b}{2 a})^{2} - (\frac{b}{2 a})^{2} + \frac{c}{a} (x + \frac{b}{2 a})^{2} (x + \frac{b}{2 a})^{2} (x + \frac{b}{2 a})^{2} x + \frac{b}{2 a} x + \frac{b}{2 a} x + \frac{b}{2 a} x x = 0 = 0 = 0 = \frac{b ^{2}}{4 a ^{2}} - \frac{c}{a} = \frac{b ^{2}}{4 a ^{2}} - \frac{4 a c}{4 a ^{2}} = \frac{b ^{2} - 4 a c}{4 a ^{2}} = \pm \frac{b ^{2} - 4 a c}{4 a ^{2}} = \frac{\pm b ^{2} - 4 a c}{4 a ^{2}} = \frac{\pm b ^{2} - 4 a c}{2 a} = \frac{- b}{2 a} + \frac{\pm b ^{2} - 4 a c}{2 a} = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Explorer

RDB

Algebra

Laws of indices

Rationalising the denominator

Quadratics

Quadratic formula

Discriminant

Hidden quadratics

Polynomials

Factor theorem

Proofs

Quadratic formula

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