An equation containing x⁴, x² and a constant—but no odd power of x—is often a quadratic in disguise. Let u = x². Then x⁴ = u², so the quartic equation becomes an ordinary quadratic in u. Solve that quadratic, reject negative u-values when real x is required, and return to x by taking both positive and negative square roots.
For example, x⁴ − 5x² + 4 = 0 becomes u² − 5u + 4 = 0, so u = 1 or u = 4. Since u = x², the complete real solution set is x = −2, −1, 1, 2. This guide explains missing roots, repeated roots, zero roots, interval inequalities and why a quadratic with real u-roots may still give no real x-solutions.
What makes a quartic biquadratic?
A quartic polynomial has degree four. A biquadratic equation has the special form ax⁴ + bx² + c = 0, where a ≠ 0 and no x³ or x term appears.
The powers can be viewed as (x²)², x² and 1. That pattern is exactly the pattern of an ordinary quadratic in a new quantity u = x².
This is a structural recognition skill. Instead of applying general quartic techniques, which are far more complicated, we reuse familiar factorisation or the quadratic formula after a suitable substitution.
Not every quartic is biquadratic. The expression x⁴ + x³ − 5x² + 4 includes an x³ term, so setting u = x² does not rewrite its entire equation as a polynomial in u. Check the powers before choosing the substitution.
Why u = x² has a domain restriction
For real x, x² ≥ 0. Therefore the new variable u = x² must be non-negative. This fact is essential when the transformed quadratic is solved.
A positive u-value produces two distinct real inputs, x = √u and x = −√u. The value u = 0 produces only one distinct input, x = 0.
A negative u-value cannot equal x² for any real x. It must be rejected when the question asks for real solutions, even if it is a perfectly valid root of the transformed quadratic.
The substitution is helpful precisely because it reduces the algebraic degree, but the new variable represents a restricted expression. Returning to the original variable and its permitted values is part of the solution, not an optional afterthought.
Full worked example: four real solutions
Solve x⁴ − 5x² + 4 = 0. Introduce u = x², so x⁴ = u². The equation becomes u² − 5u + 4 = 0.
Factorise: (u − 1)(u − 4) = 0. Thus u = 1 or u = 4. Both are positive and can equal squares of real numbers.
For u = 1, solve x² = 1, giving x = −1 or 1. For u = 4, solve x² = 4, giving x = −2 or 2.
The four distinct real answers are −2, −1, 1 and 2. A student who reports only u = 1 and 4 has not returned to x, and a student who lists only 1 and 2 has lost the negative square-root branches.
Check every real answer in the original polynomial
The original expression can be factorised directly as (x² − 1)(x² − 4). Thus a root must satisfy x² = 1 or x² = 4.
The values −1 and 1 both give x² = 1, so the first factor vanishes. The values −2 and 2 both give x² = 4, so the second factor vanishes.
Substituting an x-value into the original quartic is a more meaningful final check than substituting u into the transformed quadratic alone. It confirms that the return from u to x was correct.
The factorisation also makes the complete set visible without expanding a fourth-degree expression at every candidate. Both routes agree because the substitution was algebraically legitimate and the u-domain restriction was respected.
The difference between a real u-root and a real x-root
Consider x⁴ + 5x² + 4 = 0. Setting u = x² gives u² + 5u + 4 = 0, or (u + 1)(u + 4) = 0.
The transformed quadratic has two distinct real roots, u = −1 and u = −4. Yet neither can equal x² for real x.
Therefore the original quartic has no real solutions. The existence of two real u-roots does not guarantee even one real x-root.
This is a crucial difference from solving an ordinary quadratic in x. The substituted variable u is constrained to the non-negative half-line, and negative transformed roots are inadmissible rather than ordinary real solutions of the starting equation.
Why a discriminant check alone is insufficient
The transformed quadratic u² + 5u + 4 has discriminant 25 − 16 = 9, which is positive. This confirms that it has two real u-roots.
But the roots are both negative, so the original equation x⁴ + 5x² + 4 = 0 has no real x-solution. Using the discriminant alone would give an incorrect conclusion about the quartic.
To count real x-roots, first determine the real roots of the quadratic in u, then inspect which are positive, zero or negative. Each positive root contributes a pair of real x-values, and zero contributes one.
The sign of the roots matters just as much as their reality. This is a useful connection to parameter questions asking for positive quadratic roots.
A transformed zero gives one input, not two
Solve x⁴ − 4x² = 0. The substitution u = x² gives u² − 4u = u(u − 4) = 0.
The two u-values are 0 and 4. From u = 0, x² = 0 gives just x = 0. From u = 4, x² = 4 gives x = −2 or 2.
The distinct real solutions are x = −2, 0 and 2. Writing x = ±0 does not give two distinct answers; both signs refer to the same number.
The original factorisation x²(x − 2)(x + 2) shows that zero is a repeated root of the quartic. Multiplicity and the number of distinct input values should not be confused.
A repeated positive u-root gives two distinct real values
Solve x⁴ − 2x² + 1 = 0. The transformed quadratic is u² − 2u + 1 = (u − 1)² = 0.
There is one repeated u-root, u = 1. Returning to x gives x² = 1 and therefore x = −1 or 1.
The original polynomial is (x² − 1)² = (x − 1)²(x + 1)². Both ±1 are repeated roots, but there are only two distinct real values.
Do not report four different answers merely because the original polynomial has degree four or because each factor is squared. Degree counts multiplicity over the complex numbers, while a requested list of distinct real solutions contains each number once.
A simple non-monic biquadratic
Solve 2x⁴ − 5x² + 2 = 0. Let u = x² ≥ 0. The transformed equation is 2u² − 5u + 2 = 0.
Factorise as (2u − 1)(u − 2) = 0. Therefore u = 1/2 or u = 2. Both are positive.
From x² = 1/2, the solutions are x = ±1/√2, equivalently ±√2/2. From x² = 2, the solutions are x = ±√2.
There are four distinct real solutions. Keeping surd forms exact makes it easy to verify the substitutions and avoids round-off error in the final answers.
What if the transformed quadratic does not factor neatly?
Consider x⁴ − 3x² + 1 = 0. Substitution yields u² − 3u + 1 = 0.
Use the quadratic formula: u = [3 ± √(9 − 4)]/2 = (3 ± √5)/2. Both values are positive because 3 > √5.
Therefore x = ±√[(3 + √5)/2] or x = ±√[(3 − √5)/2]. These four exact real solutions need not be converted into decimals.
The method is unchanged: solve the quadratic in u, check admissibility, then take both square-root signs for each positive u. Factorisation is convenient when available but is not necessary for the substitution to work.
Even symmetry explains the paired roots
A biquadratic polynomial f(x) = ax⁴ + bx² + c satisfies f(−x) = f(x), because all powers of x are even.
Its graph is symmetric about the y-axis. Consequently, whenever a non-zero real root r exists, −r is also a root.
That is why the example x⁴ − 5x² + 4 has roots in symmetric pairs ±1 and ±2. Zero, when it is a root, is its own negative and therefore appears only once as a distinct input.
Graph symmetry is a useful check on a solution list. If a student reports x = 2 but not x = −2 for a biquadratic equation, something has been missed unless an additional original-domain restriction excludes the negative value.
A graphical interpretation of the substitution
The map u = x² sends every real x to a non-negative u. A quadratic equation in u identifies the values of x² for which the original quartic output is zero.
A positive u-value corresponds to two symmetric x-coordinates. The value u = 0 corresponds to the centre x = 0. Negative u-values have no real x-coordinates.
This creates a visual way to predict the number of real roots before taking square roots. Two distinct positive u-roots typically give four distinct real x-roots. One positive and one negative root give two; two negative roots give none.
Remember that repeated u-roots require careful counting, and extra restrictions such as x ≥ 0 can remove one member of each ± pair.
A domain restriction can remove the negative roots
Suppose x⁴ − 5x² + 4 = 0 is solved only for x ≥ 0. The unrestricted real roots are −2, −1, 1 and 2.
The stated domain x ≥ 0 removes −2 and −1, leaving x = 1 and x = 2. Both satisfy the original quartic and the supplied domain.
This illustrates why the ± step is necessary before final filtering. The equation alone has paired roots; the problem’s domain determines which members are admissible final answers.
A student should not delete negative roots merely because the substitution involved a square. Negative x-values can have positive squares. Delete them only when an actual problem restriction makes them invalid.
The parameter version can change the number of roots
Consider x⁴ − (k + 1)x² + k = 0, where k is real. With u = x², the transformed quadratic factorises as (u − 1)(u − k) = 0.
Thus x² = 1 or x² = k. The first condition always gives real roots x = −1 and x = 1.
If k > 0 and k ≠ 1, the second condition adds two new real roots, ±√k, giving four distinct real roots. If k = 1, these coincide with ±1, leaving only two distinct roots.
If k = 0, x² = 0 adds x = 0, so there are three distinct real roots. If k < 0, no real x can satisfy x² = k, leaving only ±1. Parameter cases must be handled explicitly rather than assuming a fixed root count.
An inequality requires regions, not just equality roots
The substitution also helps solve quartic inequalities. Consider x⁴ − 5x² + 4 ≤ 0. Set u = x² and factorise (u − 1)(u − 4) ≤ 0.
The transformed quadratic is non-positive when 1 ≤ u ≤ 4. Since u = x², we require 1 ≤ x² ≤ 4.
That means the absolute value of x lies between 1 and 2 inclusive. The solution is −2 ≤ x ≤ −1 or 1 ≤ x ≤ 2.
In interval notation, the answer is [−2, −1] ∪ [1, 2]. Simply writing 1 ≤ x ≤ 4 confuses u with x and gives the wrong region.
A strict quartic inequality excludes its roots
Now solve x⁴ − 5x² + 4 < 0. The transformed inequality becomes (u − 1)(u − 4) < 0, which holds for 1 < u < 4.
Therefore 1 < x² < 4, giving −2 < x < −1 or 1 < x < 2.
The endpoints are excluded because the quartic equals zero at ±1 and ±2, while the question requires a strictly negative value.
The change from ≤ to < affects only endpoint inclusion here; the interior negative regions remain the same. A quick even-symmetry sketch of the quartic provides a useful check that two separated intervals should appear.
When the quartic is non-negative
For x⁴ − 5x² + 4 ≥ 0, the transformed quadratic (u − 1)(u − 4) is non-negative for u ≤ 1 or u ≥ 4.
But u = x² also requires u ≥ 0. Thus the admissible u-regions are 0 ≤ u ≤ 1 or u ≥ 4.
Translate them back: x² ≤ 1 means −1 ≤ x ≤ 1, while x² ≥ 4 means x ≤ −2 or x ≥ 2.
The complete answer is (−∞, −2] ∪ [−1, 1] ∪ [2, ∞). This includes x = 0, for which the quartic equals 4. It is the complement of the negative intervals, with the equality roots appropriately retained.
A useful number-line picture for the quartic
The polynomial f(x) = x⁴ − 5x² + 4 factors as (x − 2)(x − 1)(x + 1)(x + 2). Its four simple real roots are ordered −2, −1, 1 and 2.
The sign is positive for x < −2, negative between −2 and −1, positive between −1 and 1, negative between 1 and 2, and positive for x > 2.
This alternating pattern follows from the four simple factors and a positive leading coefficient. It agrees with the substituted-u sign reasoning.
A student can choose whichever method is clearer: first solve in u and translate intervals, or factor completely in x and use a sign chart. Both should give the same inequality solution. Comparing them helps verify the reasoning rather than merely the arithmetic.
Why the quartic can have a local maximum in the middle
The function f(x) = x⁴ − 5x² + 4 is even and rises to positive infinity as |x| becomes large. Its central value is f(0) = 4.
Differentiate to obtain f′(x) = 4x³ − 10x = 2x(2x² − 5). Stationary inputs occur at x = 0 and x = ±√(5/2).
The second derivative is f″(x) = 12x² − 10. At x = 0 it equals −10, so the origin is a local maximum. At x = ±√(5/2), it is positive, giving two local minima.
This explains the W-shaped graph and the two separate negative intervals. The calculus is useful enrichment but not necessary for solving the equation by substitution.
How to avoid losing the ± square-root branches
When u = 4, the statement x² = 4 has two real solutions: x = 2 and x = −2. Writing x = √4 = 2 alone loses one.
Similarly, for u = 1/2, x = ±√(1/2). The square-root symbol √ itself denotes the non-negative principal root, but solving x² = u requires both signs when u > 0.
The exception is u = 0, which produces just x = 0. It is not two distinct roots ±0.
A clear student solution writes “for each positive u, x = ±√u” before calculating the values. This makes the return from the substituted variable explicit and reduces forgotten branches.
Don’t use substitution on the wrong structure
The equation x⁴ + 2x³ − 5x² + 4 = 0 is not biquadratic because it includes the odd power x³. If u = x², the term x³ cannot be rewritten as a polynomial in u alone over arbitrary real x.
Trying to force the substitution would leave a mixture of u and x and would not produce a genuine quadratic in one variable. Other factorisation or polynomial techniques may be needed.
The same warning applies when odd powers appear in coefficients that do not cancel. The recognisable biquadratic structure is powers x⁴, x² and the constant term only.
Before starting, scan the expression for odd powers. Method selection is an important part of Additional Mathematics, particularly in mixed-topic examination questions where the technique is not named.
A first-error checklist for exam solutions
If an answer is incomplete, find the earliest issue. Did the student recognise x⁴ as (x²)²? Did they introduce u = x² ≥ 0? Did they solve the quadratic accurately? Did they reject negative u? Did they list both signs for positive u? Did they respect any original restriction on x?
These are distinct mistakes. A child who factorises in u correctly but reports u-values as x-values needs a return-substitution repair. A child who lists negative u-values as valid x² needs a domain explanation.
A student who misses a negative x-root needs practice solving x² = c, not a completely new explanation of quartic polynomials.
Assign a changed example that targets the first error, then return to a mixed paper to confirm the learner can recognise the structure without being told to use substitution.
Six practice questions with checked answers
Question one: solve x⁴ − 5x² + 4 = 0. Answer: x = −2, −1, 1, 2.
Question two: solve x⁴ + 5x² + 4 = 0 for real x. Answer: no real solutions.
Question three: solve x⁴ − 4x² = 0. Answer: x = −2, 0, 2.
Question four: solve 2x⁴ − 5x² + 2 = 0. Answer: x = ±√2 or x = ±√2/2.
Question five: solve x⁴ − 5x² + 4 ≤ 0. Answer: [−2, −1] ∪ [1, 2].
Question six: solve x⁴ − 2x² + 1 = 0. Answer: x = −1 or 1; each is repeated.
For every question, show the substituted quadratic and explain how the sign of each u-root determines the permitted x-values.
The shortest valid exam method
Write u = x², so u ≥ 0 and x⁴ = u². Rewrite the equation entirely in u.
Solve the resulting quadratic by factorisation, completing the square or the quadratic formula. Discard any negative u-root when seeking real x, and convert each non-negative u-root back to x.
For u > 0, include both signs. For u = 0, include x = 0 once. Then check the original equation and the problem’s stated domain.
For inequalities, solve the sign condition in u and translate the entire allowed u-interval back into x, rather than taking square roots of just the endpoints in an arbitrary order. The method remains compact and exact.
Syllabus suitability and further learning
Biquadratic substitution connects indices, quadratics, factorisation, graph symmetry and domain restrictions. Some quartic applications may be enrichment rather than a specifically assessed topic, depending on the student’s syllabus and teacher’s sequence.
Families should refer to the applicable SEAB information for 2026 O-Level Additional Mathematics 4049, 2027 SEC G3 K341 or 2027 SEC G2 K232 as appropriate.
Companion reading includes How to Spot a Hidden Quadratic in an Exponential Equation, How to Find Parameter Values for Two Positive Quadratic Roots and How to Disprove a False Algebraic Rule with One Counterexample. Visit the Additional Mathematics archive or enquire about eduKate Singapore small-group tutorials near Sixth Avenue MRT.
The lasting skill is recognising a familiar quadratic structure hidden inside a higher power and preserving the domain information when returning to the original variable. Properly taught kids shine a bright light into the future.