Solving f(x) = 0 Numerically — Bisection and Newton's Method
You can solve a quadratic with a formula. You cannot solve x³ − x − 2 = 0, or cos x = x, or the cubic that comes out of a Ksp or equilibrium problem, with any formula you would want to memorise. For those you use a numerical method: you make a guess, improve it, and repeat until the answer stops changing. Two methods cover almost every case a Class 11–12 or entrance-exam student meets — bisection and Newton–Raphson. This guide shows both, step by step, with the arithmetic written out.
The idea behind both methods
Both methods answer the same question: for which value of x does the function f(x) become zero? That value is called a root. Bisection trades speed for safety — it can never fail, but it is slow. Newton's method trades safety for speed — it converges very fast when it works, and it can fail completely when it does not.
Method 1 — Bisection
Bisection needs one thing to start: two values a and b where the function has opposite signs. If f(a) is negative and f(b) is positive and the function is continuous, the curve must cross zero somewhere between them.
Step 2: midpoint m = (a + b) / 2
Step 3: if f(a) × f(m) < 0 the root is in [a, m], so replace b with m; otherwise replace a with m
Step 4: repeat. After n steps the interval width is (b − a) / 2ⁿ
Worked example 1 — bisection on x³ − x − 2 = 0
Take f(x) = x³ − x − 2 on the interval [1, 2].
f(1) = 1 − 1 − 2 = −2 (negative)
f(2) = 8 − 2 − 2 = +4 (positive)
Signs are opposite, so a root lies between 1 and 2.
| Step | Interval [a, b] | Midpoint m | f(m) | New interval |
|---|---|---|---|---|
| 1 | [1, 2] | 1.5 | 3.375 − 1.5 − 2 = −0.125 | [1.5, 2] |
| 2 | [1.5, 2] | 1.75 | 5.359375 − 1.75 − 2 = +1.609375 | [1.5, 1.75] |
| 3 | [1.5, 1.75] | 1.625 | 4.291016 − 1.625 − 2 = +0.666016 | [1.5, 1.625] |
| 4 | [1.5, 1.625] | 1.5625 | 3.814697 − 1.5625 − 2 = +0.252197 | [1.5, 1.5625] |
| 5 | [1.5, 1.5625] | 1.53125 | 3.590363 − 1.53125 − 2 = +0.059113 | [1.5, 1.53125] |
After five steps the root is trapped between 1.5 and 1.53125. The true root is 1.5214 (to 4 decimals). Notice the pattern: every step halves the uncertainty, and nothing can go wrong.
How many steps do you need? The starting width here is 1, so after n steps the width is 1/2ⁿ. For an answer correct to 4 decimal places you need 1/2ⁿ < 10⁻⁴, that is 2ⁿ > 10 000. Since 2¹³ = 8192 and 2¹⁴ = 16 384, you need 14 steps. Safe, but slow — which is exactly why the next method exists.
Method 2 — Newton–Raphson
Newton's method uses the tangent line. Stand at your current guess, slide down the tangent until it hits the x-axis, and take that point as the next guess.
Here f′(x) is the derivative — the slope of the curve at that point. You need one starting guess, not an interval.
Worked example 2 — Newton on the same equation
f(x) = x³ − x − 2, so f′(x) = 3x² − 1. Start at x₀ = 1.5.
Iteration 1:
f(1.5) = 3.375 − 1.5 − 2 = −0.125
f′(1.5) = 3(2.25) − 1 = 6.75 − 1 = 5.75
x₁ = 1.5 − (−0.125 / 5.75) = 1.5 + 0.0217391 = 1.5217391
Iteration 2:
x₁² = 2.3156900, x₁³ = 3.5238760
f(x₁) = 3.5238760 − 1.5217391 − 2 = +0.0021369
f′(x₁) = 3(2.3156900) − 1 = 6.9470700 − 1 = 5.9470700
x₂ = 1.5217391 − (0.0021369 / 5.9470700) = 1.5217391 − 0.0003593 =
1.5213798
The true root is 1.5213797. Two iterations of Newton beat fourteen steps of bisection. Roughly speaking, the number of correct digits doubles each time.
Worked example 3 — square roots without a √ key
To find √5, solve f(x) = x² − 5 = 0 with f′(x) = 2x. Substituting into Newton's formula and simplifying gives the classic averaging rule:
x₀ = 2 → x₁ = ½(2 + 2.5) = 2.25
x₁ = 2.25 → 5 / 2.25 = 2.2222222 → x₂ = ½(2.25 + 2.2222222) = 2.2361111
x₂ = 2.2361111 → 5 / 2.2361111 = 2.2360249 → x₃ = ½(2.2361111 + 2.2360249) =
2.2360680
√5 = 2.2360680 to seven decimal places. Three lines of arithmetic. This is the same idea behind the mental square-root shortcuts — you guess, then average the guess with the number divided by the guess.
When Newton's method fails
This is the part textbooks skip and examiners like. Newton is not guaranteed.
A cycle. Take f(x) = x³ − 2x + 2 with
x₀ = 0.
f(0) = 2, f′(0) = −2 → x₁ = 0 − (2 / −2) = 1
f(1) = 1 − 2 + 2 = 1, f′(1) = 3 − 2 = 1 → x₂ = 1 − (1 / 1) = 0
The iteration bounces 0 → 1 → 0 → 1 forever and never reaches the root. Bisection on a
correctly chosen interval would have found it.
Newton also breaks down if f′(xn) is zero or nearly zero — the tangent is almost horizontal and throws the next guess far away — and it may converge to a different root than the one you wanted if the starting guess is poor.
Choosing between them
| Feature | Bisection | Newton–Raphson |
|---|---|---|
| What you must supply | Two points with opposite signs | One starting guess and the derivative |
| Guaranteed to converge? | Yes, if f is continuous and signs differ | No |
| Speed | Slow — one extra binary digit per step | Fast — correct digits roughly double per step |
| Needs calculus? | No | Yes, you must differentiate f |
| Best used for | A safe first bracket, or badly behaved functions | Polishing a good guess to high accuracy |
In practice the two are used together: a few bisection steps to get close, then Newton to finish. That combination is what most software equation solvers do internally.
Common mistakes
- Starting bisection without a sign change. If f(a) and f(b) have the same sign there may be no root — or two roots — in that interval. Always evaluate both ends first.
- Differentiating wrongly in Newton's formula. A wrong f′ does not give a wrong answer slowly; it usually gives no answer at all. Check the derivative before iterating.
- Rounding at every step. Keep 6–7 digits during the iterations and round only the final answer. Rounding early destroys exactly the fast convergence you are paying for.
- Stopping on the wrong test. "x stopped changing" is not the same as "f(x) is nearly zero". For a steep function check |f(x)|; for a flat function check the change in x. Safest is to check both.
- Radians vs degrees. For equations such as cos x = x, the derivative of cos x is −sin x only in radians. Set the calculator to radian mode.
Where this appears in exams
| Exam / class | Typical use |
|---|---|
| CBSE/ICSE Class 11–12 maths | Meaning of a root, tangent and derivative, approximate solutions |
| JEE Main / Advanced | Existence of a root by sign change (intermediate value idea), tangent approximation |
| Class 11–12 chemistry numericals | Cubic equations from equilibrium and solubility-product problems |
| IIT-JAM / GATE / CSIR-NET | Numerical methods and iterative solutions in physical chemistry and mathematics papers |
Always confirm the syllabus and marking scheme in the current official notification for your exam — treat the table above as a map of where the idea is used, not as a weightage claim.
Check your root instantly. Type the function and the calculator's equation solver finds where f(x) = 0 numerically — useful for confirming the answer you reached by hand, or for the cubics that come out of equilibrium problems.
Open the Equation Solver f(x) = 0 →Struggling with the maths behind chemistry numericals in Class 11–12? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India — details at abcchemistry.in.