Exams · Russian state exam, profile level

No. 15 Inequalities

An inequality from part 2: quadratic, with fractions, exponential or logarithmic. The answer is an interval or a union of intervals.

How to solve it

Quadratic inequalities

A quadratic inequality has a quadratic on the left: $x^2 - 5x + 6 \le 0$. Its graph is a parabola, and the answer is easiest to read straight off the picture: where the parabola is above the axis and where it is below.

Step by step

Move everything to the left so that the right side is zero.
Find the roots of the quadratic by solving $ax^2 + bx + c = 0$. At the roots the parabola crosses the axis.
Sketch the parabola: it opens upwards when $a > 0$ and downwards when $a < 0$.
For $>$ or $\ge$ take the parts where the parabola is above the axis; for $<$ or $\le$, where it is below.
Include the roots only when the inequality is not strict.

The leading coefficient. When it is positive, the parabola opens upwards: it is below the axis between the roots and above it outside. For $a < 0$ it is the other way round.

The roots of the quadratic, $x_1 < x_2$.

Example: $x^2 - 5x + 6 \le 0$. The roots are $2$ and $3$, the parabola opens upwards, and we need the points below the axis together with the roots: $[2, 3]$.

Common mistakes

Forgetting that a negative $a$ turns the parabola upside down, and taking the wrong part.
Mixing up «between the roots» and «outside». A quick check helps: substitute one number from your answer.
Not knowing what to do when $D < 0$. Then the quadratic has the same sign everywhere, and the answer is either every number or the empty set.

Example

Inequalities with fractions

In an inequality with fractions the unknown is in a denominator: $\frac{x - 1}{x + 2} \le 0$. You may not multiply by the denominator: its sign is unknown, and the inequality could flip. The method of intervals works instead.

Step by step

Move everything to one side and bring it to a single fraction.
Find the zeros of the numerator and of the denominator and mark them on the number line. Zeros of the denominator are always excluded (open circles): you cannot divide by zero.
These points cut the line into intervals. On each interval the fraction keeps its sign; find it by substituting any number from the interval.
Take the intervals with the sign you need. Include the zeros of the numerator only for a non-strict inequality.

The zero of the numerator: here the fraction equals zero. It is included for a non-strict sign.

The zero of the denominator: here the fraction is not defined. It is never included.

Example: $\frac{x - 1}{x + 2} \le 0$. Mark $-2$ (open) and $1$ (filled). Right of $1$ the fraction is positive, between $-2$ and $1$ negative, left of $-2$ positive. Answer: $(-2, 1]$.

Common mistakes

Multiplying both sides by the denominator without knowing its sign.
Including a zero of the denominator, where the fraction does not exist.
Alternating the signs without checking a single point. The sign need not change, for example at a factor that is squared.

Example

Exponential inequalities

An exponential inequality such as $2^{3x + 1} > \frac{1}{4}$ starts like an equation: write both sides as powers of one base. Then look at the base itself, whether it is greater or less than one.

Step by step

Write both sides as powers of one base: $\frac{1}{4} = 2^{-2}$.
If the base is greater than $1$, compare the exponents with the same sign: $2^{3x + 1} > 2^{-2}$ means $3x + 1 > -2$.
If the base is between $0$ and $1$, the sign flips when you pass to the exponents.
Solve the inequality for the exponents and write the answer as an interval.

The base. For $a > 1$ the function $a^x$ increases, so a larger power has a larger exponent. For $0 < a < 1$ it decreases, and everything is reversed.

Example: $\left(\frac{1}{3}\right)^{x - 1} \ge \frac{1}{9}$. Here $\frac{1}{9} = \left(\frac{1}{3}\right)^2$, the base is less than $1$, the sign flips: $x - 1 \le 2$, so $x \le 3$.

Common mistakes

Not flipping the sign when the base is less than one.
Comparing the exponents of powers with different bases.
Looking for a solution of something like $2^x > -5$, which holds for every $x$: a power of a positive number is always positive.

Example

Logarithmic inequalities

A logarithmic inequality, for example $\log_2(x - 3) < 2$, is like an exponential one: the base decides again. One more condition comes in, and without it the answer is wrong: the domain.

Step by step

Write down the domain: everything under a logarithm must be positive.
Write the right side as a logarithm with the same base: $2 = \log_2 4$.
If the base is greater than $1$, the sign stays when you pass to the arguments; if it is between $0$ and $1$, it flips.
Solve the inequality for the arguments and intersect the result with the domain.

The base. For $a > 1$ the logarithm increases and the sign stays; for $0 < a < 1$ it decreases and the sign flips. The conditions $> 0$ are the domain.

Example: $\log_2(x - 3) < 2$. Domain: $x > 3$. Then $2 = \log_2 4$, the base is greater than $1$: $x - 3 < 4$, $x < 7$. With the domain: $(3, 7)$.

Common mistakes

Forgetting the domain, so that the answer contains numbers where the logarithm does not exist.
Not flipping the sign for a base less than one.
Writing $\log_2 x < 3 \iff x < 8$ and losing the left end: in fact $0 < x < 8$.