Algebra

In Algebra mode every letter is a variable. Quantus simplifies, factors and solves, and shows each rule it uses. Switch to it in the top bar.

Simplify

(a+b)^2 - 2ab


SYMBOLIC	$(a + b)^{2} - 2 \cdot a \cdot b$
STEP	$a^{2} + 2 \cdot a \cdot b + b^{2} - 2a \cdot b$ — Expand binomial power using Square of a sum
STEP	$a^{2} + b^{2}$ — Combine like terms

Powers, roots, logarithms and the trigonometric identities are simplified too:

Input	Result	Rule
`a^3 \cdot a^4`	$a^{7}$	same base: add the exponents
`\sqrt{x^2}`	$\left\\|x\right\\|$	take powers out of the root
`\ln(e^3)`	$3$
`\sin^2 x + \cos^2 x`	$1$	Pythagorean identity

Factor

2x^2 - 8

2 \cdot (x - 2)(x + 2)

Write as a binomial: Difference of squares. A quadratic without a binomial form is written by completing the square: x^2 - 6x + 3 becomes $(x - 3)^{2} - 6$ . The settings Binomial formulas and Completing the square decide which form you get, see Settings.

Fractions and their domain

\frac{x^2 - 1}{x + 1}


SYMBOLIC	$\frac{x^{2} - 1}{x + 1}$
STEP	$\frac{(x - 1)(x + 1)}{x + 1}$ — Factor numerator
STEP	$x - 1$ — Cancel common factor $x + 1$
DOMAIN	$x \neq -1$

When a factor is cancelled, the value that made it zero is kept under DOMAIN, so the result says exactly where it equals the input. Sums of fractions are put over a common denominator:

\frac{1}{x} + \frac{1}{x+1}


STEP	$\frac{1}{x} \cdot \frac{x + 1}{x + 1} + \frac{1}{x + 1} \cdot \frac{x}{x}$ — Extend each fraction
STEP	$\frac{x + 1}{x(x + 1)} + \frac{x}{x(x + 1)}$ — Multiply the fractions
STEP	$\frac{x + 1 + x}{x(x + 1)}$ — Add the numerators
STEP	$\frac{2x + 1}{x(x + 1)}$
DOMAIN	$x \neq -1, 0$

Identities that hold only under a condition, such as $\sqrt{a}\sqrt{b} = \sqrt{ab}$ for $a, b \geq 0$ , are applied with the condition added to the domain. Set Identities with a condition to Only always valid to skip them instead.

Solve

Write an equation with one unknown:

Input	Result
`x^2 - 5x + 6 = 0`	$x = 2 \lor x = 3$
`\frac{x+1}{x-2} = 3`	$x = 3{,}5$
`\|x - 1\| = 2`	$x = -1 \lor x = 3$
`\sqrt{x+3} = x - 3`	$x = 6$ (squaring would also give $x = 1$ , which does not fit)
`x^2 + 1 = 0`	This equation has no real solution. — try Imaginary mode

Systems of linear equations

Write several equations separated by ;, or as \begin{cases} … \end{cases} in the LaTeX field (open it with \):

Input	Result
`2x + y = 5; x - y = 1`	$x = 2 \land y = 1$
`\begin{cases} x + y + z = 6 \\ x - y = 0 \\ 2z = 6 \end{cases}`	$x = 1{,}5 \land y = 1{,}5 \land z = 3$
`x + y = 1; 2x + 2y = 2`	$x = 1 - y \land y \in \mathbb{R}$
`x + y = 1; x + y = 2`	The system has no solution: its equations contradict each other.

The unknowns are the letters without a value, in alphabetical order. The system is solved exactly by Gauss–Jordan elimination: at Standard detail the steps show the augmented matrix and its reduced form, at Detailed every row operation ( $\text{II} - 2 \cdot \text{I}$ ). With infinitely many solutions the unknowns without a pivot stay free and the others are given in terms of them. Coefficients may be fractions, $\pi$ or stored variables (k_1 x + y = 7); the unknowns themselves must appear linearly.

When nothing happens

An expression that is already as simple as Quantus can make it is shown as typed, with a note:

The variable x has no value, so the expression is shown as typed.

Simplify​

Factor​

Fractions and their domain​

Solve​

Systems of linear equations​

When nothing happens​

See also​