Equations

Linear, quadratic, fractional, with absolute values, exponential and logarithmic.

How to solve it

Linear equations

In a linear equation the unknown appears only to the first power: $3x + 5 = x - 7$. It has one root, and a few moves find it.

Step by step

Open any brackets.
Move the $x$ terms to the left and the numbers to the right. A term changes its sign as it crosses the «=»: $3x + 5 = x - 7$ becomes $3x - x = -7 - 5$.
Collect like terms: $2x = -12$.
Divide both sides by the number in front of $x$: $x = -6$.
Check by substituting: the left side is $3 \cdot (-6) + 5 = -13$, the right side is $-6 - 7 = -13$. They match.

The number in front of $x$, its coefficient. You may divide by it only when $a \ne 0$.

Whatever is left on the right after moving and collecting terms.

If $x$ cancels out and you get a false statement like $0 = 5$, there are no roots. If you get $0 = 0$, every number is a root.

Common mistakes

Moving a term to the other side and keeping its sign.
Losing a minus when dividing: $-2x = 8$ gives $x = -4$, not $4$.
Opening a bracket with a minus in front and changing only the first sign. Correct: $-(x - 3) = -x + 3$.

Example

Quadratic equations

A quadratic equation looks like $ax^2 + bx + c = 0$ with $a \ne 0$. It has at most two roots, and one number, the discriminant, tells you how many there really are.

Step by step

Move everything to the left so that the right side is zero. Write down $a$, $b$ and $c$ with their signs.
If $c = 0$, take $x$ out of the brackets: $x(ax + b) = 0$, so the roots are $0$ and $-\frac{b}{a}$. If $b = 0$, solve for $x^2$ and take the square root.
Otherwise compute the discriminant $D = b^2 - 4ac$.
If $D > 0$ there are two roots, if $D = 0$ one, if $D < 0$ no real roots.
Check with Vieta's formulas: the roots add up to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$.

The coefficient of $x$ with its sign. In $x^2 - 5x + 6 = 0$ it is $b = -5$, so $-b = 5$.

The discriminant: it decides how many roots there are.

The coefficient of $x^2$.

Example: $x^2 - 5x + 6 = 0$, $D = 25 - 24 = 1$, $x = \frac{5 \pm 1}{2}$, so $x = 2$ and $x = 3$.

Common mistakes

Dropping the sign of $b$: in $x^2 - 5x + 6$ the coefficient is $b = -5$, and $b^2 = 25$.
Computing the discriminant before the right side is zero. Move everything to one side first.
Dividing both sides by $x$ and losing the root $x = 0$. Take $x$ out of the brackets instead.

Example

Equations with fractions

In an equation with fractions the unknown sits in a denominator: $\frac{x^2 - 1}{x - 1} = 0$. You cannot divide by zero, and the whole solution starts from that rule.

Step by step

Find the values of $x$ that make a denominator zero. They are forbidden; the rest of the numbers form the domain of the equation.
Move everything to one side and bring it to a common denominator, so that you have a single fraction.
A fraction is zero when its numerator is zero. Solve «numerator $= 0$».
Throw away the roots that were forbidden in the first step.

The numerator: set it equal to zero.

The denominator: it must not be zero, so such roots are dropped.

Example: in $\frac{x^2 - 1}{x - 1} = 0$ the numerator gives $x = 1$ and $x = -1$, but $x = 1$ makes the denominator zero. Answer: $x = -1$.

Common mistakes

Keeping a root that makes a denominator zero. This is the most common mistake.
Cancelling an expression with $x$ in it and forgetting that it could not be zero.
Multiplying only some of the terms by the missing factor when finding the common denominator.

Example

Equations with absolute values

The absolute value $|a|$ is the distance from $a$ to zero, so it is never negative: $|3| = 3$, $|-3| = 3$. An equation with an absolute value almost always splits into two ordinary equations.

Step by step

If the right side is negative, there are no roots: an absolute value is never negative.
The equation $|f(x)| = c$ with $c > 0$ splits into $f(x) = c$ and $f(x) = -c$. Solve both.
The equation $|f(x)| = |g(x)|$ also gives two: $f(x) = g(x)$ and $f(x) = -g(x)$.
Put the roots of both equations into the answer.

The expression inside the bars.

The number on the right. The rule holds for $c \ge 0$; for a negative $c$ there are no roots.

Example: $|2x - 1| = 5$. The equation $2x - 1 = 5$ gives $x = 3$, and $2x - 1 = -5$ gives $x = -2$. Answer: $-2$ and $3$.

Common mistakes

Solving only $f(x) = c$ and losing the second root.
Looking for roots when the right side is negative, although there are none.
In $|x - 3| = |x + 1|$, forgetting that the two expressions may be opposite: that gives the second equation $x - 3 = -(x + 1)$.

Example

Exponential equations

In an exponential equation the unknown is in the exponent: $2^{x + 1} = 8$. The main trick is to write both sides as powers of the same number.

Step by step

Write both sides as powers of one base: $8 = 2^3$, $\frac{1}{9} = 3^{-2}$, $\sqrt{5} = 5^{1/2}$.
When the bases are equal, the exponents are equal: $2^{x + 1} = 2^3$ means $x + 1 = 3$.
Solve the new equation: here $x = 2$.
When one base will not do, as in $2^x = 5$, the answer is a logarithm: $x = \log_2 5$.

The base: the same positive number, not equal to $1$.

The exponent on the left.

The exponent on the right. The function $a^x$ takes each of its values exactly once, so equal powers mean equal exponents.

Example: $\left(\frac{1}{3}\right)^{x} = 9$. Here $\frac{1}{3} = 3^{-1}$ and $9 = 3^2$, so $3^{-x} = 3^2$ and $x = -2$.

Common mistakes

Equating exponents when the bases differ: $2^x = 3^x$ cannot be solved that way.
Getting the sign of a fraction wrong: $\frac{1}{8} = 2^{-3}$, not $2^3$.
Mixing up the rules: $(a^m)^n = a^{mn}$, while $a^m \cdot a^n = a^{m + n}$.

Example

Logarithmic equations

The logarithm $\log_a b$ answers a question: to what power must $a$ be raised to give $b$? For instance, $\log_2 8 = 3$ because $2^3 = 8$. A logarithmic equation is almost always solved with this definition.

Step by step

Write down the domain: whatever is under a logarithm must be positive, and the base must be positive and not equal to $1$.
If the equation is $\log_a f(x) = c$, the definition gives $f(x) = a^c$.
If it is $\log_a f(x) = \log_a g(x)$, set the arguments equal: $f(x) = g(x)$.
Solve the new equation and check every root against the domain.

The base: $a > 0$ and $a \ne 1$.

The argument, the number under the logarithm. It is always positive.

The exponent to which the base is raised.

Example: $\log_2(x - 1) = 3$, so $x - 1 = 2^3 = 8$ and $x = 9$. Domain check: $9 - 1 = 8 > 0$.

Common mistakes

Skipping the domain check and keeping a root that puts a negative number or zero under the logarithm.
Raising the wrong number to the wrong power: $\log_3 x = 2$ gives $x = 3^2 = 9$, not $x = 2^3$.
Getting lost when the right side is negative: $\log_3 x = -2$ gives $x = 3^{-2} = \frac{1}{9}$. The root exists, it is just a fraction.