If an equation can be written au² + bu + c = 0 for some expression u = f(x), the quadratic methods apply to it. Solving for u is only half the work: every value of u must be sent back through f to produce the values of x, and some of them produce none.
Core ideas
The pattern to look for, the renaming that exploits it, and the step everyone forgets.
x⁴ and x², x and √x, x⁻² and x⁻¹, x⁶ and x³ are common signals because the larger power is the square of the smaller. The more general pattern is a[f(x)]² + b[f(x)] + c: the repeated object may be a whole expression such as x² - 3, not merely a power.
The substitution renames part of the equation so that a known method applies. The question was about x, so the last step is always the return journey, and an answer left in terms of u is half an answer.
If u = x² then u cannot be negative, and if u = √x then u cannot be negative either. A negative solution for u is discarded, so two values of u need not give four values of x.
The material
The shape, the substitution, and the return journey.
A quartic has no formula in this course, and Polynomials and Factoring's rational root search is a general tool that can miss irrational roots entirely. But a quartic with no odd powers is not really a quartic: it is a quadratic wearing x² where a quadratic wears x, and renaming that part is enough to finish it exactly.
An equation is in quadratic form if it can be written au² + bu + c = 0 with a ≠ 0, where u = f(x) is some expression in x. The test is that the same inner expression appears to the first and second powers, with no other dependence on x; doubled exponents are one common clue, not the definition.
The substitution is legal because it is a renaming in both directions. Every x satisfying the original gives a u = x² satisfying the quadratic, and every u satisfying the quadratic gives back exactly the x with x² = u — two of them, one, or none, depending on the sign of u. Nothing has been done to the equation; part of it has been given a shorter name.
With u = x²: u² - 3u - 4 = (u - 4)(u + 1) = 0, so u = 4 or u = -1. Back-substituting, x² = 4 gives x = ±2, while x² = -1 has no real solution, since a square of a real number is never negative.
So this quartic has two real solutions, not four. The count of u values and the count of x values are different numbers, and only the second answers the question.
Over the complex numbers of “Complex Numbers”, “Operations with Complex Numbers” and “Complex Roots of Quadratics” the story is different: x² = -1 gives x = ±i, so the same quartic has four complex solutions, 2, -2, i and -i. Which answer is wanted depends on which number system the question is asked in, and that is worth stating in the answer.
Put u = √x, so u² = x and the equation is u² - 5u + 6 = 0, giving u = 2 or u = 3. Then √x = 2 gives x = 4, and √x = 3 gives x = 9. Both check: 4 - 10 + 6 = 0 and 9 - 15 + 6 = 0.
The domain is x ≥ 0 throughout, since √x is defined nowhere else, and both answers respect it.
The restriction earns its keep in x + √x - 6 = 0. There u² + u - 6 = (u + 3)(u - 2) = 0 gives u = 2 or u = -3, but √x is never negative, so u = -3 is discarded before it is squared. Squaring it anyway would produce x = 9, and 9 + 3 - 6 = 6, not 0 — an extraneous solution of exactly the kind Rational and Radical Expressions warned about.
Put u = x⁻¹ = 1/x, so u² = x⁻² and u² - u - 6 = (u - 3)(u + 2) = 0, giving u = 3 or u = -2. Then 1/x = 3 gives x = 1/3 and 1/x = -2 gives x = -1/2. Here a negative u is perfectly acceptable — the restriction that matters for this substitution is x ≠ 0, not a sign.
The inner expression need not be a power. In (x² - 3)² - 2(x² - 3) - 8 = 0, put u = x² - 3: then u² - 2u - 8 = (u - 4)(u + 2) = 0, so u = 4 or u = -2, giving x² = 7 and x² = 1. The four solutions are ±√7 and ±1. Expanding the brackets first would have produced the same quartic the long way round.
One manipulation to avoid: reducing the degree by dividing by a power of x. x⁴ = 4x² becomes x²(x² - 4) = 0 by factoring, with solutions 0, 2 and -2; dividing both sides by x² returns x² = 4 and silently loses x = 0. Dividing by something that might be zero deletes solutions, in this course and in every later one.
A last check worth making: Polynomials and Factoring fixed that a polynomial of degree n has at most n real solutions. A quartic that comes back with five, or a solution that fails when substituted, means the return journey went wrong — usually by squaring a rejected u or by forgetting a ±.
Interactive
Substitute u = x^2, solve, substitute back
Method
Four steps, and the fourth is the one the whole lesson exists to insist on.
x⁶ - 9x³ + 8 qualifies because x⁶ = (x³)²; (x² - 3)² - 2(x² - 3) - 8 qualifies because the entire bracket repeats. x⁴ - 9x³ + 8 has neither pattern, and no one-variable quadratic substitution will make it one.
Write u = x² or u = √x explicitly, and next to it any condition it forces: u ≥ 0 for both of those, x ≠ 0 for u = 1/x. Recording it now is what makes step four quick.
Factor, complete the square, or use the formula — whichever of “Solving by Factoring”, “The Square Root Property”, “Completing the Square” and “The Quadratic Formula” fits. Nothing about this step is new.
For each u, solve f(x) = u for x. Discard the values of u the restriction forbids, and substitute every survivor into the original equation.
The exponents are 4 and 2, and 4 is twice 2. That is the whole reason a substitution is available here.
Two values of u and four values of x. Stopping at u = 4 and u = 9 is the error this lesson exists to prevent: u was a name introduced for convenience, and the question was about x.
Both checks used the original quartic rather than the quadratic in u, which is the only way to test the substitution itself. A degree 4 equation has at most four real solutions, and this one has four, so there is nothing left to find.
For a faded return journey, solve 2x⁴ - 5x² + 2 = 0 over the reals. The supplied substitution is u = x² with u ≥ 0. Factor the quadratic in u, send both permitted values back through x² = u and check one root from each pair in the original quartic. Compare with x = ±√2 and x = ±√2/2; returning only the two positive values means the square root property lost its ±.
Solve 2x⁴ - 5x² + 2 = 0 and state how many real solutions it has and why. Two lines of setup — u = x² and u ≥ 0 — make the rest routine; skipping them is what turns a four-solution problem into a two-solution answer.
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