2026-09-29 11 min

Quartic Equation Calculator

Solve the biquadratic x⁴ + bx² + c = 0 by t = x² substitution — up to four roots with honest scope notes. Free quartic solver.

Calculator

Biquadratic Coefficients

Coefficient b (x²)

Constant c

Results update live as you type

Status:x⁴ + -5x² + 4 = 0-2, -1, 1, 24/4 real

Quartic Summary

Equationx⁴ + -5x² + 4 = 0

t-quadratict² + -5t + 4 = 0, Δ = 9

Real roots-2, -1, 1, 2

NatureFour distinct real roots

Results

Optimal

Real Roots

-2, -1, 1, 2

Unfolded from t = x²

Real Count

4 of up to 4

How many real x values

Nature

Four distinct real roots

t-sign analysis outcome

t Values

t₁ = 4, t₂ = 1

Quadratic solutions in t

t-Discriminant

b² − 4c of the t-quadratic

The t-quadratic is solved exactly; each x-root is verified by substitution into x⁴ + bx² + c within 1e-9.

Visualizations

t Values vs Roots

4t₁

1t₂

4roots

Real Root Yield

4 / 4 real100%

Formula

Biquadratics hide a quadratic: squaring the variable twice means one substitution halves the degree, and square roots unfold the answers.

t = x² → t² + bt + c = 0 → x = ±√t for t ≥ 0

Alternative Form

Scope note: general quartics (with x³, x terms) need Ferrari — this page covers biquadratic form exactly

Variables

Symbol	Name	Unit	Description
b	Middle coefficient	number	Of x² in the quartic
c	Constant	number	Shared by both forms
t	Substituted square	number	t = x² ≥ 0 for real x
x	Roots	numbers	±√t pairs

Explanation

Negative t values are not failures — they are complex pairs (±i√|t|) that simply do not plot on the real line.

Derivation

x⁴ + bx² + c = (x²)² + b(x²) + c. With t = x²: t² + bt + c = 0 → t = (−b ± √Δ)/2 → x = ±√t for t ≥ 0.

Step-by-Step Calculation

Follow each step to understand how the result is calculated from your inputs.

Step 1

Substitute t = x² to depress the degree.

t² + bt + c = 0

eq:t² + -5t + 4 = 0

Result: Quadratic in t formed

Step 2

Solve the t-quadratic via its discriminant.

Δ = b² − 4c

disc:9

t1:4

t2:1

Intermediate: t = (−b ± √Δ)/2

Result: t₁ = 4, t₂ = 1

Step 3

Unfold: each non-negative t gives x = ±√t.

x = ±√t

roots:-2, -1, 1, 2

Result: Roots: -2, -1, 1, 2

Step 4

Verify each root in x⁴ + bx² + c.

substitution ≈ 0?

count:4

Intermediate: f(-2)=0; f(-1)=0

Result: Four distinct real roots

Final Answer

-2, -1, 1, 2

Worked Examples

See how the calculator works with different scenarios.

Scenario

t = 1, 4 → four roots.

Inputs

b:-5

c:4

Calculation

Δ = 25 − 16 = 9 → t = 1, 4 → x = ±1, ±2

Result

−2, −1, 1, 2

Interpretation

Full house: both t positive, four symmetric roots.

Understanding Your Result

Your Result: Biquadratic solved by t-substitution with up to four roots

Typical range: 0–4 real roots depending on t signs; Δ < 0 gives none real.

What This Means

Each non-negative t unfolds into a ±√t pair; negative t values map to complex pairs off the real line.

Why It Matters

Symmetric fourth-degree models (beams, lenses, potentials) reduce exactly this way — no approximation needed.

Professional Interpretation

Sketch mentally: positive c with large negative b makes the W shape that crosses four times.

Recommendations

•Substitute t = x² first
•Judge Δ, then t signs
•Unfold ±√t pairs
•Verify every root
•State the general-quartic scope honestly

Educational Content

Introduction

Biquadratic equations lack odd powers, so t = x² halves their degree and ordinary quadratics finish the job.

Concept Explanation

Degree-halving substitution plus sign analysis: only t ≥ 0 unfolds to real x; the rest is complex.

Applications

▸Symmetric beam deflection

▸Biquadratic potentials

▸Lens aberration models

▸Vibration mode shapes

▸Area-of-area geometry

▸Quartic curve sketching

Benefits

•Exact, no iteration
•All four roots possible
•t-analysis transparent
•Symmetry visualized
•Scope stated honestly

Limitations

•Biquadratic form only (no x³ or x terms)
•Numeric coefficients
•Real-root display focus

Best Practices

•Spot the missing odd powers
•Substitute t = x²
•Solve the t-quadratic
•Keep only t ≥ 0 for reals
•Verify in the quartic

Common Misconceptions

Myth: Every quartic is biquadratic.

Truth: Only ones without x³ and x terms qualify; general quartics need Ferrari.

Myth: Negative t is an error.

Truth: It signals complex pairs — valid, just non-real.

Myth: Four roots always exist.

Truth: Up to four real; complex pairs complete the count to four total.

Myth: Substitution changes solutions.

Truth: t = x² is reversible for t ≥ 0 — equivalence preserved.

Practical Tips

•Confirm no x³/x terms first
•Write the t-quadratic out
•Judge Δ then t signs
•Pair ± roots
•Verify in x⁴ form
•Count to four total with complex
•Note W vs U shape

Common Mistakes

Applying to general quartics

x³ or x terms disqualify — say so, do not force.

Square-rooting negative t as real

Negative t gives ±i√|t|, not real x.

Forgetting the ± pair

Each positive t yields two x-roots.

Missing the t = 0 single

t = 0 gives one x (double zero).

Skipping verification

Fourth powers magnify slips — check.

Assumptions

This calculator makes the following assumptions to provide estimates:

Biquadratic form: Equation x⁴ + bx² + c = 0.

Real coefficients: b, c real.

Exact substitution: t = x² equivalence.

Principal roots: √ ≥ 0 unfolding.

Rounded display: Six decimals.

Limitations

Be aware of these limitations when using this calculator:

Biquadratic only: General quartics need Ferrari/CAS.

Numeric: No parameters.

Real display: Complex pairs described, not listed.

Illustrative plot: SVG window.

Frequently Asked Questions

Answers to common questions about this calculator.

How do I solve x⁴ − 5x² + 4 = 0?

Set t = x²: t² − 5t + 4 = 0 gives t = 1, 4, so x = ±1, ±2.

What is a biquadratic?

A quartic with only even powers (x⁴ + bx² + c) — solvable by halving the degree.

Why do negative t values give no real x?

Because x² ≥ 0 for real x; negative t would need imaginary x = ±i√|t|.

Can a biquadratic have 3 real roots?

Yes if one t is 0 (single root 0) and the other positive (± pair) — counting multiplicity it is still four.

What if Δ < 0?

No real t exists, hence no real x — two complex pairs instead.

Does this solve all quartics?

No — only biquadratic form. General quartics need Ferrari method or numeric solvers.

How do I verify quartic roots?

Substitute into x⁴ + bx² + c; genuine roots give ≈ 0.

Why is the graph symmetric?

Even powers make f(−x) = f(x): mirror symmetry about the y-axis.

Related Calculators

Explore more calculators related to this topic.

Related Guides

Deepen your understanding with these educational resources.

Spotting biquadratics

Even powers only.

t-substitution

Halve the degree.

Sign analysis

t ≥ 0 unfolds.

Beyond biquadratic

When Ferrari calls.

Glossary

Technical terms used in mortgage calculations explained in plain language.

Biquadratic: Quartic with only even powers.
Substitution t = x²: Degree-halving move.
t-discriminant: Δ = b² − 4c of the inner quadratic.
Unfolding: x = ±√t recovery.
Complex pair: ±i√|t| from negative t.
W shape: Four-crossing graph.
Double zero: Repeated root (t = 0 or Δ = 0).
Ferrari method: General quartic radicals.
Even function: f(−x) = f(x) symmetry.

Authoritative References

Trusted sources and standards referenced by this calculator.

Wikipedia — Quartic function MathWorld — Quartic equation Khan Academy — Polynomial graphs Purplemath — Solving polynomials Math is Fun — Polynomials Cuemath — Quartic equations

Applicable Standards

Common Core HSA-APR — higher-degree structureNCTM Algebra — substitution strategies

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Last Updated:

2026-09-29

Version:

2.0.0

References:

License:

Free to use for educational and professional purposes.