Solving Equations (DP IB Maths: AA SL)

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Dan

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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
      • space y equals x plus k blank ⟺ x equals y minus k
    • Multiplication and division are inverses
      • space y equals k x blank ⟺ x equals y over k
    • Taking the reciprocal is a self-inverse
      • space y equals 1 over x blank ⟺ x equals 1 over y
    • Odd powers and roots are inverses
      • space y equals x to the power of n blank ⟺ x equals n-th root of y
      • space y equals x to the power of n blank ⟺ x equals y to the power of 1 over n end exponent
    • Exponentials and logarithms are inverses
      • space y equals a to the power of x blank ⟺ x equals log subscript a y
      • space y equals straight e to the power of x left right double arrow x equals ln space 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
      • space y equals x to the power of n left right double arrow x equals plus-or-minus n-th root of y
    • Modulus functions lead to positive and negative solutions
      • space y equals open vertical bar x close vertical bar left right double arrow x equals plus-or-minus y
    • Trigonometric functions lead to infinite solutions using their symmetries
      • space y equals sin x left right double arrow x equals 2 k pi plus sin to the power of negative 1 end exponent y space space or space space space x equals left parenthesis 1 plus 2 k right parenthesis pi minus sin to the power of negative 1 end exponent y
      • space y equals cos x left right double arrow x equals 2 k pi plus-or-minus cos to the power of negative 1 end exponent y
      • space y equals tan x left right double arrow x equals k pi plus tan to the power of negative 1 end exponent 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 plus 1 equals 3 has one solution x equals 2
      • left parenthesis x plus 1 right parenthesis squared equals 3 to the power of 2 end exponent has two solutions x equals 2 and x equals negative 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 to the power of f left parenthesis x right parenthesis end exponent cross times a to the power of g left parenthesis x right parenthesis end exponent equals a to the power of f left parenthesis x right parenthesis plus g left parenthesis x right parenthesis end exponent
      • a to the power of f left parenthesis x right parenthesis end exponent over a to the power of g left parenthesis x right parenthesis end exponent equals a to the power of f left parenthesis x stretchy right parenthesis minus g left parenthesis x stretchy right parenthesis end exponent
      • open parentheses a to the power of f left parenthesis x right parenthesis end exponent close parentheses to the power of g left parenthesis x right parenthesis end exponent equals a to the power of f open parentheses x close parentheses cross times g left parenthesis x right parenthesis end exponent
      • a to the power of f open parentheses x close parentheses end exponent equals straight e to the power of f open parentheses x close parentheses ln space a end exponent
    • For logarithms use
      • log subscript a invisible function application f left parenthesis x right parenthesis plus log subscript a invisible function application g left parenthesis x right parenthesis equals log subscript a invisible function application open parentheses f open parentheses x close parentheses cross times g open parentheses x close parentheses close parentheses
      • log subscript a invisible function application f left parenthesis x right parenthesis minus log subscript a invisible function application g left parenthesis x right parenthesis equals log subscript a invisible function application open parentheses fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction close parentheses
      • n log subscript a invisible function application f left parenthesis x right parenthesis equals log subscript a invisible function application stretchy left parenthesis f open parentheses x close parentheses stretchy right parenthesis to the power of 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 squared plus 3 y minus 4 equals 0
      • 2 x to the power of 4 plus 3 x squared minus 4 equals 0 using space y equals x squared
      • 2 x plus 3 square root of x minus 4 equals 0 using space y equals square root of x
      • 2 over x to the power of 6 plus 3 over x cubed minus 4 equals 0 using space y equals 1 over x cubed
      • 2 straight e to the power of 2 x end exponent plus 3 straight e to the power of x minus 4 equals 0 using space y equals straight e to the power of x 
      • 2 cross times 25 to the power of x plus 3 cross times 5 to the power of x minus 4 equals 0 using space y equals 5 to the power of x
      • 2 to the power of 2 x plus 1 end exponent plus 3 cross times 2 to the power of x minus 4 equals 0 using space y equals 2 to the power of x
      • 2 open parentheses x cubed minus 1 close parentheses squared plus 3 open parentheses x cubed minus 1 close parentheses minus 4 equals 0 using space y equals x cubed minus 1
  • To solve:
    • Make the substitution space y equals f left parenthesis x right parenthesis
    • Solve the quadratic equation a y squared plus b y plus c equals 0 to get y1 & y2
    • Solve space f left parenthesis x right parenthesis equals y subscript 1 and space f left parenthesis x right parenthesis equals y subscript 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: left parenthesis x plus 1 right parenthesis left parenthesis 2 x minus 1 right parenthesis equals 3 left parenthesis x plus 1 right parenthesis
      • If you divide both sides by left parenthesis x plus 1 right parenthesis you get 2 x minus 1 equals 3 which gives x equals 2
      • However x equals negative 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 plus 1 equals 0
      • One where the equation has been divided by the expression: 2 x minus 1 equals 3
    • Make the equation equal zero and factorise
      • left parenthesis x plus 1 right parenthesis left parenthesis 2 x minus 1 right parenthesis minus 3 left parenthesis x plus 1 right parenthesis equals 0
      • left parenthesis x plus 1 right parenthesis left parenthesis 2 x minus 1 minus 3 right parenthesis equals 0 which gives left parenthesis x plus 1 right parenthesis left parenthesis 2 x minus 4 right parenthesis equals 0
      • Set each factor equal to zero and solve: x plus 1 equals 0 and 2 x minus 4 equals 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 plus ln open parentheses x minus 1 close parentheses equals 5 it would be incorrect to write straight e to the power of ln x end exponent plus straight e to the power of ln open parentheses x minus 1 close parentheses end exponent equals straight e to the power of 5 or x plus left parenthesis x minus 1 right parenthesis equals straight e to the power of 5
    • Instead it would be correct to write straight e to the power of ln x plus ln open parentheses x minus 1 close parentheses end exponent equals straight e to the power of 5 and then simplify from there

Worked example

Find the exact solutions for the following equations:

a)
5 minus 2 log subscript 4 x equals 0.

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

b)
x equals square root of x plus 2 end root.

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

c)
straight e to the power of 2 x end exponent minus 4 straight e to the power of x minus 5 equals 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 space f left parenthesis x right parenthesis equals g left parenthesis x right parenthesis
    • One method is to draw the graphs space y equals f left parenthesis x right parenthesis and space y equals g left parenthesis x right parenthesis
      • The solutions are the x-coordinates of the points of intersection
    • Another method is to draw the graph space y equals f left parenthesis x right parenthesis minus g left parenthesis x right parenthesis or space y equals g left parenthesis x right parenthesis minus f left parenthesis x right parenthesis
      • 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 to the power of 5 minus x plus 1 equals 0
    • Equations involving different types of functions
      • straight e to the power of x equals x squared

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

a)
Sketch the graph y equals straight e to the power of x minus x squared.

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

b)
Hence find the solution to straight e to the power of x equals x squared.

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

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Dan

Author: Dan

Expertise: Maths

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.