☰ Contents · Algebra

Systems of second-degree inequalities

Lessons 16 · 1 lessons · Sh.A. Alimov, A.R. Xalmukhamedov, M.A. Mirzaakhmedov. Algebra: textbook for Grade 9 of general secondary schools. Revised 4th edition. Tashkent: “O‘qituvchi”, 2019
16

Systems of second-degree inequalities in one unknown

Textbook: pp. 77–79
GoalSolve systems of inequalities that include a quadratic inequality, inequalities with absolute value, and domain problems.
New words
system of inequalities · tengsizliklar sistemasiintersection of solution sets · yechimlar kesishmasiinequality with absolute value · modulli tengsizlikequivalent inequalities · tengkuchli tengsizliklar
Explanation

A solution of a system of inequalities is any x satisfying all the inequalities at once. So we solve each inequality separately (a quadratic one by the graph or the interval method), mark the solutions on one number line and take their common part. Inequalities with absolute value reduce to systems: for a > 0, |f(x)| < a is equivalent to −a < f(x) < a, i.e. a system of two inequalities; |f(x)| > a is equivalent to f(x) < −a or f(x) > a. Finding a domain also gives a system: each expression under a square root must be non-negative, and a denominator must not be zero. Write the answer from the common part on the number line; if there is no common part, the system has no solution.

Worked examples
x² − 4x + 3 > 0 and 2x + 4 ≥ 0. The first: x < 1 or x > 3. The second: x ≥ −2. Common part: −2 ≤ x < 1 or x > 3.
|x² − 5| < 4 ⇔ −4 < x² − 5 < 4. This is the system x² − 1 > 0 (x < −1 or x > 1) and x² − 9 < 0 (−3 < x < 3). Answer: −3 < x < −1 or 1 < x < 3.
Class activity

“Find the common part”: two students mark the solutions of two inequalities on one number line with different coloured pencils; the part coloured by both is the solution of the system.

Practice
1
Solve the system x² − 9 ≤ 0 and x − 1 > 0.
2
Find the domain of y = √(x² − 5x + 6) + √(4 − x).
3
Solve |x² − 1| < 3.
4
Why is |f| < a equivalent to −a < f < a (a > 0)?