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

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

    • $y=x+3 ⟺x=y-3$

  • Multiplication and division are inverses

    • $y=3x ⟺x=\frac{y}{3}$

  • Taking the reciprocal is a self-inverse

    • $y=\frac{1}{x} ⟺x=\frac{1}{y}$

  • Odd powers and roots are inverses

    • $y=x^3 ⟺x=\sqrt[3]{y}=y^{\frac{1}{3}}$

  • Exponentials and logarithms are inverses

    • $y=3^x ⟺x={\log }_{3}y$

    • $y={\mathrm{e}}^x\iff x=\ln y$

• 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

    • $y=x^4\iff x=\pm \sqrt[4]{y}$

  • Modulus functions lead to positive and negative solutions

    • $y=\left|x\right|\iff x=\pm y$

  • Trigonometric functions lead to infinite solutions using their symmetries

    • $y=\sin x\iff x=2k\pi +{\sin }^{-1}y\mathrm{or}x=(1+2k)\pi -{\sin }^{-1}y$

    • $y=\cos x\iff x=2k\pi \pm {\cos }^{-1}y$

    • $y=\tan x\iff x=k\pi +{\tan }^{-1}y$

• 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

    • $a^{f\left(x\right)}\times a^{g\left(x\right)}=a^{f\left(x\right)+g\left(x\right)}$

    • $\frac{a^{f\left(x\right)}}{a^{g\left(x\right)}}=a^{f\left(x\right)-g\left(x\right)}$

    • ${\left(a^{f\left(x\right)}\right)}^{g\left(x\right)}=a^{f\left(x\right)\times g\left(x\right)}$

    • $a^{f\left(x\right)}={\mathrm{e}}^{f\left(x\right)\ln a}$

  • For logarithms use

    • ${\log }_{a}f\left(x\right)+{\log }_{a}g\left(x\right)={\log }_a\left(f\left(x\right)\times g\left(x\right)\right)$

    • ${\log }_{a}f\left(x\right)-{\log }_{a}g\left(x\right)={\log }_a\left(\frac{f\left(x\right)}{g\left(x\right)}\right)$

    • $n{\log }_{a}f\left(x\right)={\log }_a{\left(f\left(x\right)\right)}^n$

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 $2y^2+3y-4=0$

    • $2x^4+3x^2-4=0$ using $y=x^2$

    • $2x+3\sqrt{x}-4=0$ using $y=\sqrt{x}$

    • $\frac{2}{x^6}+\frac{3}{x^3}-4=0$ using $y=\frac{1}{x^3}$

    • $2{\mathrm{e}}^{2x}+3{\mathrm{e}}^x-4=0$ using $y={\mathrm{e}}^x$ 

    • $2\times {25}^x+3\times 5^x-4=0$ using $y=5^x$

    • $2^{2x+1}+3\times 2^x-4=0$ using $y=2^x$

    • $2{\left(x^3-1\right)}^2+3\left(x^3-1\right)-4=0$ using $y=x^3-1$

• To solve:

  • Make the substitution $y=f\left(x\right)$

  • Solve the quadratic equation $a{y}^2+by+c=0$ to get y₁ & y₂

  • Solve $f\left(x\right)=y_1$ and $f\left(x\right)=y_2$

    • 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: $(x+1)(2x-1)=3(x+1)$

    • If you divide both sides by $(x+1)$ you get $2x-1=3$ 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: $2x-1=3$

  • Make the equation equal zero and factorise

    • $(x+1)(2x-1)-3(x+1)=0$

    • $(x+1)(2x-1-3)=0$ which gives $(x+1)(2x-4)=0$

    • Set each factor equal to zero and solve: $x+1=0$ and $2x-4=0$