☰ Contents · Algebra

Reviewing the Grade 8 course

Lessons 1 · 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
1

Review of the topics studied in Grade 8

Textbook: pp. 3–4
GoalRecall the main Grade 8 topics: linear function, systems, absolute value, quadratic equations, Vieta's theorem.
New words
linear function · chiziqli funksiyadiscriminant · diskriminantVieta's theorem · Viyet teoremasibiquadratic equation · bikvadrat tenglama
Explanation

In Grade 8 you worked with the linear function y = kx + b and its graph, a straight line: k is the slope and b is the ordinate where the graph meets the Oy axis. A system of two equations in two unknowns is solved by addition or by substitution. For the quadratic equation ax² + bx + c = 0 the discriminant is D = b² − 4ac: if D > 0 there are two roots, if D = 0 one root, and if D < 0 there is no real root. For the reduced equation x² + px + q = 0, Vieta's theorem says x₁ + x₂ = −p and x₁ · x₂ = q. In a biquadratic equation ax⁴ + bx² + c = 0 we substitute t = x². We need all this in the new chapter on the quadratic function and quadratic inequalities.

Worked examples
A line passes through (0, 4) and (2, 10): b = 4 (at x = 0 we get y = b), and 2k + 4 = 10 gives k = 3. The equation is y = 3x + 4.
x² − 7x + 12 = 0: D = 49 − 48 = 1, x = (7 ± 1) : 2, so x₁ = 4, x₂ = 3. Check: x₁ + x₂ = 7 = −p, x₁ · x₂ = 12 = q.
Class activity

“From roots to equation”: in pairs, one student thinks of two roots (for example 2 and −5), the other builds x² + px + q = 0 with Vieta's theorem, and then the first checks it with the discriminant.

Practice
1
Solve 2x² − 5x + 2 = 0.
2
Solve the system x + y = 7, x − y = 1.
3
Solve the inequality |x − 2| < 3.
4
Why does x² + 4 = 0 have no real root?