Equations Reducible to Quadratic Form

Some equations that look complicated are actually quadratics in disguise. An equation is “reducible to quadratic form” when a substitution transforms it into $au^2 + bu + c = 0$. Once you solve for $u$, you substitute back to find the original variable. The technique applies to fourth-degree polynomials, equations with rational exponents, and certain exponential equations. The crucial final step is checking for extraneous solutions — not every value of $u$ produces a valid value of the original variable.

Recognizing the Pattern

An equation is reducible to quadratic form when it contains:

• A term with some expression squared and the same expression to the first power
• No other powers of that expression

The general pattern:

$a[f(x)]^2 + b[f(x)] + c = 0$

Let $u = f(x)$, and you get $au^2 + bu + c = 0$.

Common substitutions:

Worked Examples

Example 1: Fourth-Degree Polynomial

Solve $x^4 - 13x^2 + 36 = 0$.

Step 1 — Identify the substitution: $u = x^2$, so $u^2 = x^4$.

$u^2 - 13u + 36 = 0$

Step 2 — Solve the quadratic:

$(u - 4)(u - 9) = 0$

$u = 4 \quad \text{or} \quad u = 9$

Step 3 — Back-substitute:

If $u = 4$: $x^2 = 4 \Rightarrow x = \pm 2$

If $u = 9$: $x^2 = 9 \Rightarrow x = \pm 3$

Step 4 — Check: Substitute each into the original:

• $x = 2$: $16 - 52 + 36 = 0$ — valid
• $x = -2$: $16 - 52 + 36 = 0$ — valid
• $x = 3$: $81 - 117 + 36 = 0$ — valid
• $x = -3$: $81 - 117 + 36 = 0$ — valid

Solutions: $x = -3, -2, 2, 3$.

Example 2: Rational Exponents

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

Step 1: Let $u = x^{1/3}$, so $u^2 = x^{2/3}$.

$u^2 - 5u + 6 = 0$

Step 2: $(u - 2)(u - 3) = 0 \Rightarrow u = 2$ or $u = 3$.

Step 3 — Back-substitute:

If $u = 2$: $x^{1/3} = 2 \Rightarrow x = 2^3 = 8$

If $u = 3$: $x^{1/3} = 3 \Rightarrow x = 3^3 = 27$

Step 4 — Check:

• $x = 8$: $8^{2/3} - 5(8^{1/3}) + 6 = 4 - 10 + 6 = 0$ — valid
• $x = 27$: $27^{2/3} - 5(27^{1/3}) + 6 = 9 - 15 + 6 = 0$ — valid

Solutions: $x = 8, 27$.

Example 3: Equation with Square Roots

Solve $x - 7\sqrt{x} + 10 = 0$.

Step 1: Let $u = \sqrt{x}$, so $u^2 = x$.

$u^2 - 7u + 10 = 0$

Step 2: $(u - 2)(u - 5) = 0 \Rightarrow u = 2$ or $u = 5$.

Step 3 — Back-substitute:

If $u = 2$: $\sqrt{x} = 2 \Rightarrow x = 4$

If $u = 5$: $\sqrt{x} = 5 \Rightarrow x = 25$

Step 4 — Check:

• $x = 4$: $4 - 7(2) + 10 = 4 - 14 + 10 = 0$ — valid
• $x = 25$: $25 - 7(5) + 10 = 25 - 35 + 10 = 0$ — valid

Solutions: $x = 4, 25$.

Example 4: Exponential Equation

Solve $e^{2x} - 7e^x + 12 = 0$.

Step 1: Let $u = e^x$, so $u^2 = e^{2x}$.

$u^2 - 7u + 12 = 0$

Step 2: $(u - 3)(u - 4) = 0 \Rightarrow u = 3$ or $u = 4$.

Step 3 — Back-substitute:

If $u = 3$: $e^x = 3 \Rightarrow x = \ln 3$

If $u = 4$: $e^x = 4 \Rightarrow x = \ln 4$

Step 4 — Check: Since $e^x > 0$ for all $x$, both $u = 3 > 0$ and $u = 4 > 0$ are valid.

Solutions: $x = \ln 3 \approx 1.099$ and $x = \ln 4 \approx 1.386$.

Example 5: Composite Expression Substitution

Solve $(x^2 + 3x)^2 - 14(x^2 + 3x) + 40 = 0$.

Step 1: Let $u = x^2 + 3x$.

$u^2 - 14u + 40 = 0$

Step 2: $(u - 4)(u - 10) = 0 \Rightarrow u = 4$ or $u = 10$.

Step 3 — Back-substitute and solve each quadratic:

If $u = 4$: $x^2 + 3x = 4 \Rightarrow x^2 + 3x - 4 = 0 \Rightarrow (x+4)(x-1) = 0$

$x = -4$ or $x = 1$

If $u = 10$: $x^2 + 3x = 10 \Rightarrow x^2 + 3x - 10 = 0 \Rightarrow (x+5)(x-2) = 0$

$x = -5$ or $x = 2$

Step 4 — Check all four values in the original equation:

• $x = -4$: $(-4)^2 + 3(-4) = 4$, so $(4)^2 - 14(4) + 40 = 16 - 56 + 40 = 0$ — valid
• $x = 1$: $(1)^2 + 3(1) = 4$, so $0$ — valid
• $x = -5$: $(-5)^2 + 3(-5) = 10$, so $(10)^2 - 14(10) + 40 = 0$ — valid
• $x = 2$: $(2)^2 + 3(2) = 10$, so $0$ — valid

Solutions: $x = -5, -4, 1, 2$.

Example 6: When Extraneous Solutions Appear

Solve $x^{2/3} + x^{1/3} - 2 = 0$.

Step 1: Let $u = x^{1/3}$.

$u^2 + u - 2 = 0 \Rightarrow (u + 2)(u - 1) = 0$

$u = -2$ or $u = 1$.

Step 2 — Back-substitute:

If $u = -2$: $x^{1/3} = -2 \Rightarrow x = (-2)^3 = -8$

If $u = 1$: $x^{1/3} = 1 \Rightarrow x = 1$

Step 3 — Check:

• $x = -8$: $(-8)^{2/3} + (-8)^{1/3} - 2 = 4 + (-2) - 2 = 0$ — valid
• $x = 1$: $1 + 1 - 2 = 0$ — valid

Both are valid here. But consider a similar equation where $u$ must be non-negative (like when $u = \sqrt{x}$) — then a negative $u$ value would be extraneous.

When to Expect Extraneous Solutions

Extraneous solutions arise when:

• Square root substitution ($u = \sqrt{x}$): $u$ must be $\ge 0$, so negative $u$-values are discarded
• Even-root substitution ($u = x^{1/4}$): same constraint — $u \ge 0$
• Exponential substitution ($u = e^x$ or $u = 2^x$): $u$ must be $> 0$, so zero or negative $u$-values are extraneous
• Domain restrictions in the original equation (denominators, logarithms)

Always check by substituting back into the original equation.

Real-World Application: Electrical Engineering

In AC circuit analysis, the impedance $Z$ of a circuit with both capacitive and inductive elements can lead to equations of the form:

$\omega^4 - 10\omega^2 + 9 = 0$

where $\omega$ is the angular frequency (always positive in practice).

Let $u = \omega^2$:

$u^2 - 10u + 9 = 0 \Rightarrow (u-1)(u-9) = 0$

$u = 1$ or $u = 9$.

$\omega^2 = 1 \Rightarrow \omega = 1$ rad/s (since $\omega > 0$)

$\omega^2 = 9 \Rightarrow \omega = 3$ rad/s

These are the resonant frequencies of the circuit — the frequencies at which certain circuit behaviors (like maximum current or voltage) occur. The negative roots $\omega = -1$ and $\omega = -3$ are mathematically valid but physically meaningless.

The Complete Process Summary

1. Spot the pattern — look for an expression and its square
2. Define $u$ as the simpler expression
3. Rewrite the equation as a quadratic in $u$
4. Solve the quadratic (factoring, formula, or completing the square)
5. Back-substitute — solve for the original variable from each $u$-value
6. Check every solution in the original equation — discard extraneous ones

Common Mistakes

1. Forgetting to back-substitute. Solving for $u$ is only half the problem — you need the original variable.
2. Not checking for extraneous solutions. Especially with radicals and exponentials, not all $u$-values yield valid original solutions.
3. Missing the substitution. Practice recognizing the pattern: if you see $x^4$ and $x^2$ (but no $x^3$ or $x$), think $u = x^2$.
4. Forgetting $\pm$ when taking square roots. If $x^2 = 9$, then $x = 3$ AND $x = -3$.
5. Wrong exponent arithmetic. Remember: $(x^{1/3})^2 = x^{2/3}$, not $x^{2/9}$.

Practice Problems

Key Takeaways

• An equation is reducible to quadratic form when it contains an expression and its square, with no other powers
• $u$-substitution transforms the equation into a standard quadratic $au^2 + bu + c = 0$
• Common substitutions: $u = x^2$, $u = x^{1/3}$, $u = \sqrt{x}$, $u = e^x$, or $u = (\text{expression in } x)$
• After solving for $u$, you must back-substitute to find the original variable
• Always check for extraneous solutions — especially when the substitution involves radicals, even powers, or exponentials
• Domain constraints (like $\sqrt{x} \ge 0$ or $e^x > 0$) eliminate invalid $u$-values before back-substitution

Return to College Algebra for more topics in this section.