Solving Equations (DP IB Analysis & Approaches (AA)): Revision Note

Author

Dan Finlay

Last updated

11 December 2024

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Solving Equations Analytically

How can I solve equations analytically where the unknown appears only once?

These equations can be solved by rearranging
For one-to-one functions you can just apply the inverse
- Addition and subtraction are inverses
  - $y = x + k ⟺ x = y - k$
- Multiplication and division are inverses
  - $y = k x ⟺ x = \frac{y}{k}$
- Taking the reciprocal is a self-inverse
  - $y = \frac{1}{x} ⟺ x = \frac{1}{y}$
- Odd powers and roots are inverses
  - $y = x^{n} ⟺ x = \sqrt[n]{y}$
  - $y = x^{n} ⟺ x = y^{\frac{1}{n}}$
- Exponentials and logarithms are inverses
  - $y = a^{x} ⟺ x = \log_{a} y$
  - $y = e^{x} \Leftrightarrow x = \ln y$
For many-to-one functions you will need to use your knowledge of the functions to find the other solutions
- Even powers lead to positive and negative solutions
  - $y = x^{n} \Leftrightarrow x = \pm \sqrt[n]{y}$
- Modulus functions lead to positive and negative solutions
  - $y = |x| \Leftrightarrow x = \pm y$
- Trigonometric functions lead to infinite solutions using their symmetries
  - $y = \sin x \Leftrightarrow x = 2 k π + \arcsin y or x = (1 + 2 k) π - \arcsin y$
  - $y = \cos x \Leftrightarrow x = 2 k π \pm arc \cos y$
  - $y = \tan x \Leftrightarrow x = k π + arc \tan y$
Take care when you apply many-to-one functions to both sides of an equation as this can create additional solutions which are incorrect
- For example: squaring both sides
  - $x + 1 = 3$ has one solution $x = 2$
  - ${(x + 1)}^{2} = 3^{2}$ has two solutions $x = 2$ and $x = - 4$
Always check your solutions by substituting back into the original equation

How can I solve equations analytically where the unknown appears more than once?

Sometimes it is possible to simplify expressions to make the unknown appear only once
Collect all terms involving x on one side and try to simplify into one term
- For exponents use
  - $a^{f (x)} \times a^{g (x)} = a^{f (x) + g (x)}$
  - $\frac{a^{f (x)}}{a^{g (x)}} = a^{f (x) - g (x)}$
  - ${(a^{f (x)})}^{g (x)} = a^{f (x) \times g (x)}$
  - $a^{f (x)} = e^{f (x) \ln a}$
- For logarithms use
  - $\log_{a} f (x) + \log_{a} g (x) = \log_{a} (f (x) \times g (x))$
  - $\log_{a} f (x) - \log_{a} g (x) = \log_{a} (\frac{f (x)}{g (x)})$
  - $n \log_{a} f (x) = \log_{a} {(f (x))}^{n}$

How can I solve equations analytically when the equation can't be simplified?

Sometimes it is not possible to simplify equations
Most of these equations cannot be solved analytically
A special case that can be solved is where the equation can be transformed into a quadratic using a substitution
- These will have three terms and involve the same type of function
Identify the suitable substitution by considering which function is a square of another
- For example: the following can be transformed into $2 y^{2} + 3 y - 4 = 0$
  - $2 x^{4} + 3 x^{2} - 4 = 0$ using $y = x^{2}$
  - $2 x + 3 \sqrt{x} - 4 = 0$ using $y = \sqrt{x}$
  - $\frac{2}{x^{6}} + \frac{3}{x^{3}} - 4 = 0$ using $y = \frac{1}{x^{3}}$
  - $2 e^{2 x} + 3 e^{x} - 4 = 0$ using $y = e^{x}$
  - $2 \times 25^{x} + 3 \times 5^{x} - 4 = 0$ using $y = 5^{x}$
  - $2^{2 x + 1} + 3 \times 2^{x} - 4 = 0$ using $y = 2^{x}$
  - $2 {(x^{3} - 1)}^{2} + 3 (x^{3} - 1) - 4 = 0$ using $y = x^{3} - 1$
To solve:
- Make the substitution $y = f (x)$
- Solve the quadratic equation $a y^{2} + b y + c = 0$ to get y₁ & y₂
- Solve $f (x) = y_{1}$ and $f (x) = y_{2}$
  - Note that some equations might have zero or several solutions

Can I divide both sides of an equation by an expression?

When dividing by an expression you must consider whether the expression could be zero
Dividing by an expression that could be zero could result in you losing solutions to the original equation
- For example: $(x + 1) (2 x - 1) = 3 (x + 1)$
  - If you divide both sides by $(x + 1)$ you get $2 x - 1 = 3$ which gives $x = 2$
  - However $x = - 1$ is also a solution to the original equation
To ensure you do not lose solutions you can:
- Split the equation into two equations
  - One where the dividing expression equals zero: $x + 1 = 0$
  - One where the equation has been divided by the expression: $2 x - 1 = 3$
- Make the equation equal zero and factorise
  - $(x + 1) (2 x - 1) - 3 (x + 1) = 0$
  - $(x + 1) (2 x - 1 - 3) = 0$ which gives $(x + 1) (2 x - 4) = 0$
  - Set each factor equal to zero and solve: $x + 1 = 0$ and $2 x - 4 = 0$

Examiner Tips and Tricks

A common mistake that students make in exams is applying functions to each term rather than to each side
- For example: Starting with the equation $\ln x + \ln (x - 1) = 5$ it would be incorrect to write $e^{\ln x} + e^{\ln (x - 1)} = e^{5}$ or $x + (x - 1) = e^{5}$
- Instead it would be correct to write $e^{\ln x + \ln (x - 1)} = e^{5}$ and then simplify from there

Worked Example

Find the exact solutions for the following equations:

a) $5 - 2 \log_{4} x = 0$ .

2-4-3-ib-aa-sl-solve-analytically-a-we-solution

b) $x = \sqrt{x + 2}$ .

2-4-3-ib-aa-sl-solve-analytically-b-we-solution

c) $e^{2 x} - 4 e^{x} - 5 = 0$ .

2-4-3-ib-aa-sl-solve-analytically-c-we-solution

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Solving Equations Graphically

How can I solve equations graphically?

To solve $f (x) = g (x)$
- One method is to draw the graphs $y = f (x)$ and $y = g (x)$
  - The solutions are the x-coordinates of the points of intersection
- Another method is to draw the graph $y = f (x) - g (x)$ or $y = g (x) - f (x)$
  - The solutions are the roots (zeros) of this graph

Why do I need to solve equations graphically?

Some equations cannot be solved analytically
- Polynomials of degree higher than 4
  - $x^{5} - x + 1 = 0$
- Equations involving different types of functions
  - $e^{x} = x^{2}$

Examiner Tips and Tricks

On a calculator paper you are allowed to solve equations using your GDC unless the question asks for an algebraic method
If your answer needs to be an exact value then you might need to solve analytically to get the exact value

Worked Example

a) Sketch the graph $y = e^{x} - x^{2}$ .

2-4-3-ib-aa-sl-solve-graphically-a-we-solution

b) Hence find the solution to $e^{x} = x^{2}$ .

2-4-3-ib-aa-sl-solve-graphically-b-we-solution

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Previous:Exponential & Logarithmic FunctionsNext:Modelling with Functions

Solving Equations (DP IB Analysis & Approaches (AA)): Revision Note

Solving Equations Analytically

How can I solve equations analytically where the unknown appears only once?

How can I solve equations analytically where the unknown appears more than once?

How can I solve equations analytically when the equation can't be simplified?

Can I divide both sides of an equation by an expression?

Solving Equations Graphically

How can I solve equations graphically?

Why do I need to solve equations graphically?

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

1. Number & Algebra

Number & Algebra Toolkit

Standard Form

Laws of Indices

Partial Fractions

Exponentials & Logs

Introduction to Logarithms

Laws of Logarithms

Solving Exponential Equations

Sequences & Series

Language of Sequences & Series

Arithmetic Sequences & Series

Geometric Sequences & Series

Applications of Sequences & Series

Compound Interest & Depreciation

Simple Proof & Reasoning

Proof

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Binomial Theorem

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Extension of The Binomial Theorem

Permutations & Combinations

Counting Principles

Permutations & Combinations

Complex Numbers

Introduction to Complex Numbers

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Further Complex Numbers

Geometry of Complex Numbers

Forms of Complex Numbers

Complex Roots of Polynomials

De Moivre's Theorem

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Systems of Linear Equations

Systems of Linear Equations

Algebraic Solutions

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