Solving Equations | DP IB Maths: AA HL Revision Notes 2021

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 Tip

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:

5 - 2 \log_{4} x = 0

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

x = \sqrt{x + 2}

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

e^{2 x} - 4 e^{x} - 5 = 0

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

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
    - This method is sometimes quicker as it involves drawing only one 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 Tip

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

Sketch the graph

y = e^{x} - x^{2}

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

Hence find the solution to

e^{x} = x^{2}

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

Solving Equations (DP IB Maths: AA HL)

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?

Examiner Tip

Worked example

Solving Equations Graphically

How can I solve equations graphically?

Why do I need to solve equations graphically?

Examiner Tip

Worked example

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Author: Dan