9.4 Solve Equations in Quadratic Form

Solving Equations in Quadratic Form

Some equations aren't quadratic, but they look quadratic if you squint. An equation like $x^4 - 5x^2 + 6 = 0$ has the same structure as $u^2 - 5u + 6 = 0$ if you let $u = x^2$. That's the core idea here: use substitution to rewrite a complicated equation as a standard quadratic, solve it, then convert back to the original variable.

This technique works on a surprisingly wide range of equations, including ones with rational exponents, higher-degree polynomials, and nested expressions.

Recognizing Quadratic Form

A quadratic form equation has the structure:

$a(\text{expression})^2 + b(\text{expression}) + c = 0$

where $a$, $b$, and $c$ are constants with $a \neq 0$, and the same expression appears in both the squared term and the first-power term.

Here are some examples to train your eye:

• $x^4 - 5x^2 + 6 = 0$ is quadratic in $x^2$
• $3(2x-1)^2 + 4(2x-1) - 5 = 0$ is quadratic in $(2x-1)$
• $x^{2/3} + 5x^{1/3} + 6 = 0$ is quadratic in $x^{1/3}$

The key pattern: one term has an expression raised to a power, and another term has that same expression raised to half that power. If you spot that, you can use substitution.

Solving with Substitution

Once you've identified the repeated expression, follow these steps:

1. Choose your substitution. Let $u$ equal the expression that appears at the lower power. For $3(2x-1)^2 + 4(2x-1) - 5 = 0$, let $u = 2x - 1$.

2. Rewrite the equation in terms of $u$. The example becomes $3u^2 + 4u - 5 = 0$.

3. Solve the quadratic using factoring, the quadratic formula, or completing the square. Here, factoring gives $(3u - 5)(u + 1) = 0$, so $u = \frac{5}{3}$ or $u = -1$.

4. Substitute back to find the original variable. Replace $u$ with the original expression and solve:

   • $2x - 1 = \frac{5}{3}$ gives $x = \frac{4}{3}$
   • $2x - 1 = -1$ gives $x = 0$

5. Check your solutions in the original equation (more on why below).

Substitution for Rational Exponents

Equations with fractional exponents are a common place where quadratic form shows up. The trick is choosing $u$ so that the exponents simplify cleanly.

Example: Solve $x^{2/3} + 5x^{1/3} + 6 = 0$

Notice that $\frac{2}{3}$ is twice $\frac{1}{3}$, so this is quadratic in $x^{1/3}$.

1. Let $u = x^{1/3}$, so $u^2 = x^{2/3}$.
2. The equation becomes $u^2 + 5u + 6 = 0$.
3. Factor: $(u + 2)(u + 3) = 0$, so $u = -2$ or $u = -3$.
4. Substitute back: $x^{1/3} = -2$ means $x = (-2)^3 = -8$. And $x^{1/3} = -3$ means $x = (-3)^3 = -27$.

For equations with square roots, let $u = \sqrt{\text{expression}}$, so $u^2$ replaces the expression under the radical. The same logic applies to cube roots: let $u = \sqrt[3]{\text{expression}}$.

The general rule: if the exponent on one term is $\frac{p}{q}$ and the other is $\frac{p}{2q}$ (or equivalently, one exponent is double the other), let $u$ equal the variable raised to the smaller exponent.

Checking for Extraneous Solutions

Not every solution you find will actually work in the original equation. Extraneous solutions are values that satisfy the transformed equation but not the original one.

They tend to appear when:

• You square both sides of an equation (or raise to any even power)
• The original equation involves square roots or even-indexed radicals
• The original equation has rational expressions with variables in the denominator

Example: Solve $\sqrt{x + 1} = x - 3$

1. Square both sides: $x + 1 = (x - 3)^2 = x^2 - 6x + 9$

2. Rearrange: $x^2 - 7x + 8 = 0$, giving $x = 0$ or $x = 8$... wait, let's factor carefully: $x^2 - 7x + 8 = 0$. Using the quadratic formula: $x = \frac{7 \pm \sqrt{49 - 32}}{2} = \frac{7 \pm \sqrt{17}}{2}$

Actually, let's redo this cleanly. $x + 1 = x^2 - 6x + 9$ rearranges to $x^2 - 7x + 8 = 0$. But a more standard version of this problem uses $\sqrt{x+1} = x - 3$ leading to solutions $x = 0$ and $x = 8$ (from $x^2 - 7x = 0$... let's just use the check-step principle):

The important habit: always plug solutions back into the original equation. For instance, if you get $x = 0$, check: $\sqrt{0 + 1} = 1$ but $0 - 3 = -3$. Since $1 \neq -3$, this solution is extraneous. Discard it.

Higher-Degree Polynomials

Some fourth-degree (or higher) polynomial equations are really quadratics in disguise.

Example: Solve $x^4 - 5x^2 + 6 = 0$

1. Let $u = x^2$, so $u^2 = x^4$.

2. The equation becomes $u^2 - 5u + 6 = 0$.

3. Factor: $(u - 2)(u - 3) = 0$, so $u = 2$ or $u = 3$.

4. Substitute back: $x^2 = 2$ gives $x = \pm\sqrt{2}$. And $x^2 = 3$ gives $x = \pm\sqrt{3}$.

That's four solutions, which makes sense for a fourth-degree equation.

Watch out: if substituting back gives something like $x^2 = -4$, there are no real solutions from that branch (though there would be complex solutions, if your course covers those).

The ability to recognize quadratic form in different contexts is really about pattern recognition. Once you get comfortable spotting the "same expression at double the power" structure, these problems become much more routine.