4.3 Solving via Substitution

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📖 Main Content

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What is the Substitution Method?

Sometimes you will be given an equation that looks complicated — it might have large powers, fractions as powers, logarithms, or exponentials. But hidden inside it is actually a quadratic equation (an equation of the form $au^2 + bu + c = 0$).

The substitution method works like this:

Once you solve the quadratic, you substitute back to find the actual values of $x$.

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The Three Steps — Always Follow These

Step 1: Spot the pattern and choose your substitution. Look at the equation and find a part that repeats or that, when squared (or manipulated), gives another term in the equation. Let $u$ equal that part.

Step 2: Rewrite the equation using $u$ and solve the quadratic. Replace the complicated part everywhere it appears. You should now have a clean quadratic: $au^2 + bu + c = 0$. Solve it by factorising or the quadratic formula.

Step 3: Substitute back and solve for $x$. Replace $u$ with whatever it was equal to (in terms of $x$), and solve the resulting equation(s) for $x$.

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Type 1 — Equations with Fractional or Large Powers

These are equations where $x$ appears with powers like $x^4$, $x^{2/3}$, $x^{4/3}$, etc.

Key idea: Notice that one power is exactly double the other. That is the signal to use substitution.

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Example 1: Solve $4x^4 - 17x^2 + 4 = 0$

Step 1 — Choose the substitution: Notice that $x^4 = (x^2)^2$. So if we let $u = x^2$, then $x^4 = u^2$.

Step 2 — Rewrite and solve:

$4u^2 - 17u + 4 = 0$

Factorise: $(4u - 1)(u - 4) = 0$

$$u = \frac{1}{4} \quad \text{or} \quad u = 4$$

Step 3 — Substitute back ($u = x^2$):

$x^2 = \frac{1}{4} \implies x = \pm\frac{1}{2}$

$x^2 = 4 \implies x = \pm 2$

✅ Final answers: $x = -2, \; -\frac{1}{2}, \; \frac{1}{2}, \; 2$