Algebra Formulas

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

solves $ax^2+bx+c=0$ for $x$, where $a \neq 0$

$\Delta = b^2-4ac$

$\Delta > 0$: two distinct real roots

$\Delta = 0$: one repeated real root

$\Delta < 0$: two complex conjugate roots

$x_1+x_2=-\frac{b}{a}$

sum of the roots

$x_1 x_2=\frac{c}{a}$

product of the roots

$a^m\cdot a^n=a^{m+n}$

$\frac{a^m}{a^n}=a^{m-n}$

$a \neq 0$

$(a^m)^n=a^{mn}$

$(ab)^n=a^n b^n$

$a^0=1$

$a \neq 0$

$a^{-n}=\frac{1}{a^n}$

$a \neq 0$

$a^{\frac{1}{n}}=\sqrt[n]{a}$

$a^{\frac{m}{n}}=\sqrt[n]{a^m}$

$\log_b(xy)=\log_b x+\log_b y$

$\log_b\left(\frac{x}{y}\right)=\log_b x-\log_b y$

$\log_b(x^n)=n\log_b x$

$\log_b b=1$

$\log_b 1=0$

$\log_b x=\frac{\log_k x}{\log_k b}$

change of base — $k$ is any valid base

$b^{\log_b x}=x$

$a^2-b^2=(a-b)(a+b)$

difference of squares

$a^2+2ab+b^2=(a+b)^2$

perfect square trinomial

$a^2-2ab+b^2=(a-b)^2$

perfect square trinomial

$a^3+b^3=(a+b)(a^2-ab+b^2)$

sum of cubes

$a^3-b^3=(a-b)(a^2+ab+b^2)$

difference of cubes

$a_n=a_1+(n-1)d$

arithmetic sequence — $n$th term

$S_n=\frac{n}{2}(a_1+a_n)$

arithmetic series — sum of the first $n$ terms

$a_n=a_1 r^{n-1}$

geometric sequence — $n$th term

$S_n=a_1\frac{1-r^n}{1-r}$

geometric series — sum of the first $n$ terms, $r \neq 1$

$S=\frac{a_1}{1-r}$

infinite geometric series — converges only when $|r| < 1$

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$

$\binom{n}{k}=\frac{n!}{k!(n-k)!}$