Solve equations reducible to quadratic form by a suitable substitution, including equations in powers, surds and exponentials

What this dot point is asking

SEAB wants you to spot when an equation that is not obviously a quadratic becomes one after a substitution, to make that substitution, solve the quadratic, and then revert carefully to find all valid solutions in the original variable. The art is choosing the right substitution and not losing or inventing solutions on the way back.

The answer

Recognising a hidden quadratic

An equation is reducible to quadratic form when one power or expression is the square of another. Tell-tale signs:

• $x^4$ and $x^2$ together (a quadratic in $x^2$),
• $x$ and $\sqrt{x}$ together (a quadratic in $\sqrt{x}$),
• $a^{2x}$ and $a^x$ together (a quadratic in $a^x$).

Making the substitution

Let $u$ stand for the simpler expression (such as $u = x^2$, $u = \sqrt{x}$ or $u = a^x$). The equation then has the form $au^2 + bu + c = 0$, which you solve by factorising or the formula.

Reverting and checking

Replace $u$ by the original expression and solve for the variable. Two cautions:

• A value of $u$ may give two values of the original variable (for $u = x^2$, each positive $u$ gives $x = \pm\sqrt{u}$).
• A value of $u$ may give none: $u = a^x$ or $u = \sqrt{x}$ can only be positive, so a negative or zero $u$ is rejected.

Why checking matters

Squaring or substituting can introduce values that do not satisfy the original equation, so always test which reverted values are genuinely valid.

Spotting the substitution from the structure

The fastest way to choose the substitution is to look for a term that is the square of another term in the equation. Whenever you see a power that is exactly double another, $x^4$ doubling $x^2$, $a^{2x}$ doubling $a^x$, or $x$ as the square of $\sqrt{x}$, let $u$ be the smaller of the pair. The equation then collapses to $au^2 + bu + c = 0$. Checking that the highest power is precisely twice the middle power before substituting confirms the equation really is reducible; if the powers are not in a $2:1$ ratio, no single substitution will make it quadratic and a different method is needed.

Counting the solutions you should expect

Knowing how many solutions to expect guards against losing some. A quadratic in $u$ gives up to two values of $u$, and each is reverted to the original variable, so the final solution count depends on the substitution: $u = x^2$ can give up to four real values of $x$ (two signs for each positive $u$), while $u = a^x$ gives at most one $x$ per valid positive $u$. So $x^4 - 13x^2 + 36 = 0$ has four roots, but $3^{2x} - 4(3^x) + 3 = 0$ has only two. Predicting the expected number of roots from the substitution is a built-in check that you have not dropped a sign or an impossible value.

Examples in context

Example 1. Trigonometric equations. An equation such as $2\sin^2\theta - 3\sin\theta + 1 = 0$ is a quadratic in $\sin\theta$; substituting $u = \sin\theta$, solving, and reverting (keeping only $-1 \leq u \leq 1$) is exactly this technique applied in trigonometry.

Example 2. Exponential modelling. Population or decay models that mix $e^{2t}$ and $e^{t}$ reduce to a quadratic in $e^{t}$, letting you solve for the time at which two effects balance without graphing.

Try this

Q1. Solve $x^4 - 5x^2 + 4 = 0$. [3 marks]

• Cue. $u = x^2$: $(u - 1)(u - 4) = 0$, so $x = \pm 1, \pm 2$.

Q2. Solve $2^{2x} - 6(2^x) + 8 = 0$. [4 marks]

• Cue. $u = 2^x$: $(u - 2)(u - 4) = 0$, so $2^x = 2$ or $4$, giving $x = 1$ or $2$.

Q3. Solve $x - 4\sqrt{x} + 3 = 0$. [3 marks]

• Cue. $u = \sqrt{x}$: $(u - 1)(u - 3) = 0$, so $x = 1$ or $x = 9$.