# Substitution to Form a Quadratic

> CIE IGCSE Additional Mathematics · 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u4-substitution-to-form-a-quadratic/

This guide teaches you to use algebraic substitution to reduce hidden quadratic equations (logarithmic, exponential, index form) to standard $ax^2+bx+c=0$ form for solving, aligned to CIE IGCSE Additional Mathematics 0606 syllabus 4.3.

**Prerequisites:** Solving quadratic equations by factorisation, completing the square or quadratic formula; Manipulating logarithmic expressions; Manipulating exponential and index expressions

## Learning objectives

- Recognise hidden quadratic equations in logarithmic, exponential and index form
- Select appropriate substitutions to reduce hidden quadratics to standard $ax^2+bx+c=0$ form
- Solve the resulting quadratic and reverse substitution to find valid solutions for the original variable
- Verify solutions and discard invalid values (e.g. negative logarithm arguments)

## Identifying Hidden Quadratics & Choosing Substitutions

A hidden quadratic is an equation where a single repeated expression appears both squared and to the first power, plus a constant term. The goal of substitution is to replace this repeated expression with a single variable (usually $y$ or $u$) to rewrite the equation as standard quadratic $ay^2 + by + c = 0$. To choose the right substitution, locate the base of the squared term in the equation.

**Appropriate Substitution** — The repeated expression that forms the base of the squared term in the original equation, replaced with a new variable to form a standard quadratic.

*Example:* For $2(\ln 5x)^2 + \ln 5x - 6 = 0$, the repeated base is $\ln 5x$, so use substitution $y = \ln 5x$.

**Worked example:** Identify the appropriate substitution for the equation $3(2^x)^2 + 2^x - 10 = 0$

1. Step 1: Locate the squared term in the equation
2. $$(2^x)^2$$
3. Step 2: Identify the base of the squared term, which is the repeated expression appearing elsewhere in the equation
4. $$2^x$$
5. Step 3: Assign this expression to a new variable. The appropriate substitution is:
6. $$y = 2^x$$

> **Exam tip:** If the question provides a substitution, use it exactly as given to avoid losing method marks, even if you think another substitution is easier.

## Substitution for Logarithmic Equations

Logarithmic hidden quadratics contain a logarithmic expression that appears both squared and linear. After substituting, solve the quadratic, then reverse the substitution to solve for $x$, remembering that $\ln k$ is only defined for $k > 0$, so discard any solutions that lead to a negative or zero argument of the logarithm.

**Worked example:** Solve $2(\ln 5x)^2 + \ln 5x - 6 = 0$, giving your answers to 3 significant figures where appropriate.

1. Step 1: Choose the substitution for the repeated logarithmic expression
2. $$y = \ln 5x$$
3. Step 2: Substitute into the original equation to form a quadratic
4. $$2y^2 + y - 6 = 0$$
5. Step 3: Factorise and solve the quadratic
6. $$(2y - 3)(y + 2) = 0 \implies y = \frac{3}{2} \text{ or } y = -2$$
7. Step 4: Reverse the substitution for each solution
8. First solution: $\ln 5x = 1.5 \implies 5x = e^{1.5} \implies x = \frac{e^{1.5}}{5} \approx 0.896$
9. Second solution: $\ln 5x = -2 \implies 5x = e^{-2} \implies x = \frac{e^{-2}}{5} \approx 0.0271$
10. Step 5: Verify both solutions give positive arguments for $\ln 5x$, so both are valid.

> **warning**
>
> Always check that your solutions for $x$ do not make the argument of a logarithm negative or zero, as these values are invalid and will be marked wrong if included in your final answer.

*Calculator:* allowed

## Substitution for Exponential Equations

Exponential hidden quadratics often contain terms of $e^x$ and $e^{-x}$, or the same base raised to $x$ and $2x$. For equations with $e^{-x}$, multiply all terms by $e^x$ first to eliminate the negative index before substituting, as $e^x > 0$ for all real $x$ so this does not change the equality.

**Worked example:** Solve $3e^x = 12 - 5e^{-x}$, giving answers to 2 decimal places.

1. Step 1: Eliminate the negative exponent by multiplying all terms by $e^x$
2. $$3e^{2x} = 12e^x - 5$$
3. Step 2: Rearrange to standard quadratic form
4. $$3e^{2x} -12e^x +5 =0$$
5. Step 3: Substitute $y = e^x$
6. $$3y^2 -12y +5 =0$$
7. Step 4: Solve using quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where $a=3, b=-12, c=5$
8. $$y = \frac{12 \pm \sqrt{144 - 60}}{6} = \frac{12 \pm \sqrt{84}}{6} \approx 3.5275 \text{ or } 0.4725$$
9. Step 5: Reverse substitution, discard negative $y$ as $e^x > 0$ (both solutions positive here)
10. $$e^x = 3.5275 \implies x = \ln(3.5275) \approx 1.26$$
11. $$e^x = 0.4725 \implies x = \ln(0.4725) \approx -0.75$$

> **Exam tip**
>
> For exponential equations with negative powers, multiply through by the positive power first to get a standard squared term before substitution. This is a high-frequency tested step in exams.

*Calculator:* allowed

## Substitution for Index Equations

Index equations use integer bases raised to powers, such as $2^x$, $3^{2x}$, etc. Remember that $a^{2x} = (a^x)^2$, so this is the squared term you will use for your substitution, even if it is written as a different base (e.g. $4^x = (2^2)^x = (2^x)^2$).

**Worked example:** Solve $4^x - 6(2^x) + 8 = 0$

1. Step 1: Rewrite $4^x$ as a power of 2 to get the same base
2. $$4^x = (2^2)^x = (2^x)^2$$
3. Step 2: Rewrite the original equation
4. $$(2^x)^2 -6(2^x) +8 =0$$
5. Step 3: Substitute $y=2^x$
6. $$y^2 -6y +8 =0$$
7. Step 4: Factorise and solve the quadratic
8. $$(y-2)(y-4)=0 \implies y=2 \text{ or } y=4$$
9. Step 5: Reverse substitution
10. $$2^x = 2 \implies x=1$$
11. $$2^x=4 \implies x=2$$

**Check your understanding**

1. What is the correct substitution for the equation $9^x - 4(3^x) +3=0$?

   - $y=x$
   - $y=3^x$
   - $y=9^x$
   - $y=x^2$

   *Why:* Correct! $9^x=(3^x)^2$, so $y=3^x$ forms the quadratic $y^2 -4y +3=0$.

*Calculator:* forbidden

## Substitution for Root and Fractional-Power Equations

Hidden quadratics also appear with roots and fractional powers, such as $\sqrt{x}$ (that is, $x^{1/2}$) or $x^{1/3}$. The key idea is that the higher power is the square of the lower one: $x = (\sqrt{x})^2$ and $x^{2/3} = (x^{1/3})^2$, so you substitute for the lower power. **Domain care is essential:** $\sqrt{x}$ is defined only for $x \geq 0$ and is itself never negative, so any negative value of $y = \sqrt{x}$ must be rejected. A cube root $x^{1/3}$, by contrast, is defined for all real $x$ and can be negative, so negative values of $y = x^{1/3}$ are kept.

**Root / Fractional-Power Substitution** — For an equation of the form $ax + b\sqrt{x} + c = 0$ use $y = \sqrt{x}$ (with $y \geq 0$); for $ax^{2/3} + bx^{1/3} + c = 0$ use $y = x^{1/3}$ (any real $y$).

*Example:* $2x - 11\sqrt{x} + 12 = 0$ becomes $2y^2 - 11y + 12 = 0$ with $y = \sqrt{x}$.

**Worked example:** Solve $2x - 11\sqrt{x} + 12 = 0$.

1. Step 1: Substitute $y = \sqrt{x}$, so $x = y^2$ and the constraint $y \geq 0$ applies.

   $$2y^2 - 11y + 12 = 0$$
2. Step 2: Factorise and solve the quadratic.

   $$(2y - 3)(y - 4) = 0 \implies y = \frac{3}{2} \text{ or } y = 4$$
3. Step 3: Both roots satisfy $y \geq 0$, so neither is rejected. (Had a root been negative, it would be discarded because $\sqrt{x}$ cannot be negative.)
4. Step 4: Reverse the substitution using $x = y^2$.

   $$y = \frac{3}{2} \implies x = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \\ y = 4 \implies x = 4^2 = 16$$
5. Step 5: Check in the original equation. For $x = \frac{9}{4}$: $2 \cdot \frac{9}{4} - 11 \cdot \frac{3}{2} + 12 = \frac{9}{2} - \frac{33}{2} + 12 = -12 + 12 = 0$. For $x = 16$: $2(16) - 11(4) + 12 = 32 - 44 + 12 = 0$. Both are valid, so $x = \frac{9}{4}$ or $x = 16$.

**Worked example:** Solve $x^{2/3} - 5x^{1/3} + 6 = 0$.

1. Step 1: Note that $x^{2/3} = (x^{1/3})^2$, so substitute $y = x^{1/3}$.

   $$y^2 - 5y + 6 = 0$$
2. Step 2: Factorise and solve the quadratic.

   $$(y - 2)(y - 3) = 0 \implies y = 2 \text{ or } y = 3$$
3. Step 3: $y = x^{1/3}$ is a cube root, which is defined for all real $x$ and may be negative, so no root is rejected on domain grounds (here both are positive anyway).
4. Step 4: Reverse the substitution by cubing, since $x = y^3$.

   $$y = 2 \implies x = 2^3 = 8 \\ y = 3 \implies x = 3^3 = 27$$
5. Step 5: Check in the original equation. For $x = 8$: $8^{2/3} - 5 \cdot 8^{1/3} + 6 = 4 - 10 + 6 = 0$. For $x = 27$: $27^{2/3} - 5 \cdot 27^{1/3} + 6 = 9 - 15 + 6 = 0$. Both are valid, so $x = 8$ or $x = 27$.

> **warning**
>
> Do not treat $\sqrt{x}$ and $x^{1/3}$ the same way. Because $\sqrt{x} \geq 0$, a negative value of $y = \sqrt{x}$ has no solution and must be discarded. But a cube root can be negative, so a negative value of $y = x^{1/3}$ is valid and gives a negative value of $x$.

> **Exam tip:** Set your substitution equal to the term with the smaller power so the larger power becomes its square: for $\sqrt{x}$ use $y = \sqrt{x}$ (then $x = y^2$); for $x^{1/3}$ use $y = x^{1/3}$ (then $x^{2/3} = y^2$).

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to reverse the substitution after solving the quadratic, giving $y$ values as final answers.
  - Why it fails: The question asks for solutions to the original equation in terms of $x$, not the substituted variable $y$.
  - Correct: Always substitute back to the original variable and solve for $x$ after solving the quadratic.
- **Wrong:** Including invalid solutions where the logarithm argument is zero or negative.
  - Why it fails: Logarithms are only defined for positive arguments, so these values are not valid real solutions.
  - Correct: After solving for $x$, substitute back into the original logarithmic expression to confirm the argument is positive, discard any invalid solutions.
- **Wrong:** Failing to eliminate negative exponents in exponential equations before substitution.
  - Why it fails: This leads to incorrect quadratic formation, often with a negative power that cannot be solved as a standard quadratic.
  - Correct: Multiply all terms by the positive version of the exponential term to eliminate negative powers before substituting.
- **Wrong:** Choosing the squared term itself as the substitution (e.g. $y = (\ln x)^2$ instead of $y = \ln x$).
  - Why it fails: This leads to a linear equation in $y$ rather than a quadratic, and you will not be able to solve correctly.
  - Correct: Choose the base of the squared term, not the squared term itself, as your substitution variable.
- **Wrong:** Discarding positive exponential solutions that are less than 1.
  - Why it fails: $e^x$ and $a^x$ for positive $a>1$ are always positive for all real $x$, so values between 0 and 1 are valid.
  - Correct: Only discard negative $y$ solutions for exponential/index equations, as these have no real solutions for $x$.

## Cheatsheet

| Equation Type | Substitution | Key Check |
| --- | --- | --- |
| Logarithmic: $a(\ln f(x))^2 + b(\ln f(x)) + c =0$ | $y = \ln f(x)$ | $f(x) > 0$ for valid solutions |
| Exponential: $a e^{2x} + b e^x + c =0$ | $y = e^x$ | Discard $y < 0$ |
| Index: $a k^{2x} + b k^x + c =0$ | $y = k^x$ | Discard $y < 0$ for positive $k$ |
| Exponential with negative power: $a e^x + b + c e^{-x}=0$ | Multiply by $e^x$ first, then $y=e^x$ | No zero division (always true for $e^x$) |
| Root / fractional power: $ax + b\sqrt{x} + c =0$ or $ax^{2/3} + bx^{1/3} + c =0$ | $y=\sqrt{x}$ (then $x=y^2$) or $y=x^{1/3}$ (then $x=y^3$) | For $\sqrt{x}$ discard $y<0$; a cube-root $y$ may be negative |

## What's next

Now that you can use substitution to form and solve hidden quadratics, you are ready to apply this skill to more complex equation types in the CIE IGCSE Additional Mathematics 0606 syllabus. This technique is frequently combined with other topics including graph sketching (to find the number of solutions) and calculus (to find stationary points of functions with hidden quadratic forms). Mastering substitution will also help you with higher-level topics like trigonometric equations in later units, where the same approach is used to reduce trigonometric expressions to quadratics.

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