Lessons 13–14 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 10, Parts 1 and 2, 1st edition. EXTREMUM PRESS, Tashkent, 2017
13
Solving equations approximately
Textbook, Part 1: pp. 74–76
GoalLocate the interval containing a root of a polynomial by a sign change and compute the root to a given accuracy by the bisection method.
New words
approximate solution · taqribiy yechimbisection · oraliqni teng ikkiga bo‘lishaccuracy · aniqlikinterval containing a root · ildiz yotgan oraliq
Explanation
The roots of many equations cannot be found by an exact formula, so they are computed approximately. Key fact: if a polynomial f(x) has values of opposite signs at the ends of [a, b], i.e. f(a) · f(b) < 0, then f(x) = 0 has at least one root inside the segment (the graph crosses the x-axis). The converse is false: a root may exist even if f(a) · f(b) > 0 (a root repeated an even number of times). In the bisection method we take a segment [a, b] with one root, find its midpoint m = (a + b)/2 and compute f(m); if f(a) · f(m) < 0 the root is in [a, m], otherwise in [m, b]. At each step the segment halves; we stop when b − a < ε and take the midpoint as the approximate root with accuracy ε.
Worked examples
f(x) = x³ + x − 3. f(1) = −1 < 0 and f(2) = 7 > 0, so there is a root in (1, 2). Midpoint 1.5: f(1.5) = 1.875 > 0 ⇒ root in (1, 1.5). Midpoint 1.25: f(1.25) ≈ 0.203 > 0 ⇒ (1, 1.25). Midpoint 1.125: f(1.125) ≈ −0.451 < 0 ⇒ (1.125, 1.25). The length is 0.125 and the root is ≈ 1.2.
For f(x) = x² − 2x + 1 we have f(0) = 1 and f(2) = 1, the product is positive, yet x = 1 is a (double) root. So f(a) · f(b) > 0 does not mean there is no root; only f(a) · f(b) < 0 guarantees a root.
Class activity
Pair game “Which side?”: one student thinks of a number from 1 to 100, the other guesses and only hears “higher” or “lower”. The fastest way is to name the middle of the interval each time; discuss how this resembles bisection.
Practice
1
Which condition guarantees a root of the polynomial f(x) in [a, b]?
f(a) · f(b) < 0 (the values at the ends have opposite signs).
2
For f(x) = x³ + 2x − 5 compute f(1) · f(2).
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3
Narrow the root of f(x) = x³ + 2x − 5 in [1, 2] by one bisection step: compute f(1.5) and name the new interval.
f(1.5) = 3.375 + 3 − 5 = 1.375 > 0 and f(1) = −2 < 0 ⇒ the new interval is [1, 1.5].
4
Why is the new interval [1, 1.5] and not [1.5, 2]?
The sign change is on [1, 1.5]: f(1) = −2 < 0, f(1.5) = 1.375 > 0. On [1.5, 2] both values are positive (1.375 and 7), so there is no guarantee.