# Solving equations: algebraic and graphical methods

> IB Mathematics: Applications and Interpretation HL · Unit 2: Functions
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u2-solving-equations-algebraic-and-graphical/

This module covers core techniques for finding roots of univariate equations, combining exact algebraic methods for solvable polynomial, rational and exponential forms, and approximate graphical methods for non-linear equations that cannot be solved exactly.

**Prerequisites:** [Basic algebraic manipulation and factorization](https://www.owlsprep.com/study/ib-math-ai-hl-u1-algebraic-manipulation/); [Graphing polynomial and non-linear functions](https://www.owlsprep.com/study/ib-math-ai-hl-u2-graphing-basic-functions/)

## Learning objectives

- Solve linear, quadratic, rational and exponential equations using exact algebraic methods
- Use graphical methods to find approximate roots of non-solvable non-linear equations
- Identify extraneous solutions introduced during algebraic manipulation
- Select the appropriate method for a given equation based on its form

## Algebraic methods for polynomial and rational equations

**Solution of an equation** — A value $x=a$ that makes the equation true, i.e. $f(a)=0$ when the equation is rearranged to equal zero

*Notation:* For $f(x) = 0$

*Example:* $x=2$ is a solution to $2x - 4 = 0$, since $2(2) - 4 = 0$

Algebraic methods produce exact roots for many common equation types. For linear equations $ax + b = 0$, the solution is always $x = -\frac{b}{a}$. For quadratic equations $ax^2 + bx + c = 0$, we can use factorization, completing the square, or the quadratic formula to find exact roots.

**Worked example:** Solve $\frac{2x}{x-1} - \frac{3}{x+1} = 2$, find all valid solutions

1. Multiply both sides by the common denominator $(x-1)(x+1)$ to eliminate fractions:
2. $$2x(x+1) - 3(x-1) = 2(x-1)(x+1)$$
3. Expand all terms:
4. $$2x^2 + 2x - 3x + 3 = 2x^2 - 2$$
5. Simplify by collecting like terms:
6. $$-x + 5 = 0 \implies x = 5$$
7. Check for extraneous solutions by substituting back into the original equation:
8. $$\text{LHS} = \frac{2(5)}{5-1} - \frac{3}{5+1} = 2.5 - 0.5 = 2 = \text{RHS}$$
9. $x=5$ is a valid solution

> **Exam tip:** Always check for extraneous solutions when you multiply by an expression containing the unknown, this is required for full marks on IB exams.

## Algebraic methods for exponential and logarithmic equations

Exponential equations with the unknown in the exponent can often be solved exactly using logarithm properties. We use the power rule $\ln(a^b) = b\ln(a)$ to bring the exponent down to a linear term, then solve for the unknown.

**Worked example:** Solve $3^{2x - 1} = 7^{x + 2}$ for $x$, give your answer to 3 decimal places

1. Take the natural logarithm of both sides:
2. $$\ln\left(3^{2x -1}\right) = \ln\left(7^{x+2}\right)$$
3. Apply the power rule for logarithms:
4. $$(2x -1)\ln 3 = (x + 2)\ln 7$$
5. Expand and collect terms with $x$ on the left-hand side:
6. $$2x \ln 3 - \ln 3 = x \ln 7 + 2\ln 7 \\ x(2\ln 3 - \ln 7) = 2\ln 7 + \ln 3$$
7. Solve for $x$ by dividing both sides:
8. $$x = \frac{2\ln 7 + \ln 3}{2\ln 3 - \ln 7}$$
9. Calculate using a calculator to get the final result:
10. $$x \approx 7.052$$

**Check your understanding**

What is the correct first step to solve $2e^{3x} = 10$?

1. What is the correct first step?

   - Divide both sides by 2 then take natural log
   - Take natural log then divide by 2
   - Set $3x = \ln(10)$
   - Divide 10 by $2e$

   *Answer:* Divide both sides by 2 then take natural log

   *Why:* You must isolate the exponential term before taking the logarithm of both sides, otherwise you cannot simplify the expression correctly.

## Graphical methods for approximate root finding

Many non-linear equations (such as $e^{-x} = x$ or $x^3 - 2x - 5 = 0$) cannot be solved exactly with algebraic methods. For these, we use graphical methods with your GDC to find accurate approximate roots.

There are two common approaches: 1) Rearrange the equation to $f(x) = 0$, then roots are the $x$-intercepts of $y=f(x)$. 2) Split the equation into $f(x) = g(x)$, then roots are the $x$-coordinates of the intersection of $y=f(x)$ and $y=g(x)$.

**Worked example:** Find an approximate solution to $e^{-x} = x$ using the intersection method

1. Split the equation into two functions to graph: $y_1 = e^{-x}$ and $y_2 = x$
2. Plot both functions on your GDC, they intersect at exactly one point for $x>0$
3. Use the GDC's 'intersect' tool to find the coordinates of the crossing point
4. Read the $x$-coordinate of the intersection to get the solution: $x \approx 0.567$
5. Verify: $e^{-0.567} \approx 0.567$, which matches the right-hand side

> **Exam tip:** Practice using the 'zero' and 'intersect' tools on your GDC before the exam, this saves significant time and avoids errors.

## Comparing algebraic and graphical methods

**Comparing methods**

Choose your method based on the form of the equation and the required output:

- **Algebraic** — Exact method for standard solvable equation types
  - Pros: Produces exact, analytic roots; No calculator required for basic cases
  - Cons: Only works for specific equation types; Can produce extraneous solutions

- **Graphical** — Approximate method for any graphed equation
  - Pros: Works for any equation you can plot; Shows the number of roots visually
  - Cons: Only gives approximate roots; Requires a graphic display calculator

> **tip**
>
> Always check the number of roots by looking at the graph, even if you solve algebraically, to confirm you haven't missed any solutions.

## Common pitfalls

- **Wrong:** Forgetting to check for extraneous solutions after solving rational or logarithmic equations
  - Why it fails: Multiplying both sides by an expression containing $x$ can introduce values that make the original denominator zero or log argument negative
  - Correct: Always substitute all obtained solutions back into the original equation to confirm they satisfy it
- **Wrong:** Missing multiple roots when only using an algebraic method
  - Why it fails: Higher-degree polynomials and non-linear equations often have more roots than the initial algebraic manipulation finds
  - Correct: Plot the function on your GDC to confirm how many roots exist before you start solving
- **Wrong:** Rounding intermediate steps when solving algebraically
  - Why it fails: Early rounding leads to inaccurate final answers, especially for exponential and logarithmic equations
  - Correct: Keep full calculator precision through all steps, only round the final answer to the required accuracy
- **Wrong:** Reporting the $y$-coordinate of an intersection as the solution
  - Why it fails: The solution to $f(x) = g(x)$ is the value of $x$ that makes the equation true
  - Correct: Always confirm you report the $x$-coordinate of the intersection as your solution

## Cheatsheet

| Equation Type | Algebraic Method | Graphical Alternative |
| --- | --- | --- |
| Linear $ax+b=0$ | $x=-b/a$ | $x$-intercept of $y=ax+b$ |
| Quadratic $ax^2+bx+c=0$ | Factorize or quadratic formula | $x$-intercepts of parabola |
| Rational with fractions | Multiply by denominator, check for extraneous roots | Intersection of LHS and RHS |
| Exponential $a^{kx+b}=c^{mx+n}$ | Take log, solve linear for $x$ | Intersection of two exponentials |
| Non-solvable non-linear | Not applicable | Find $x$-intercept or intersection |

## What's next

Solving equations is a foundational skill for all further topics in IB AI HL, from applied modelling to calculus and optimization. Mastering the choice between exact and approximate methods prepares you for real-world problem solving, where most applied equations cannot be solved exactly with algebraic techniques. This sub-topic builds directly on basic function knowledge and leads into core applied topics including systems of equations, numerical methods, and optimization of functions for real-world models.

- [Inverse functions: algebraic solution](https://www.owlsprep.com/study/ib-math-ai-hl-u2-inverse-functions-algebraic-solution/)
- [Modelling with functions](https://www.owlsprep.com/study/ib-math-ai-hl-u2-modelling-with-functions/)

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