Calculator
Example data table
| Model | Inputs | Verified solution(s) | Notes |
|---|---|---|---|
| Model 1 | √(3x + 1) = 2 | x = 1 | Constant right side fits the same model. |
| Model 2 | √(x) + √(x + 9) = 9 | x = 16 | Domain requires x ≥ 0. |
| Model 3 | √(3x + 6) = √(x + 10) | x = 2 | Linear after squaring, then domain check. |
Formula used
- √(u) = v implies u = v², but only when u ≥ 0 and v ≥ 0.
- √(u) + √(w) = e is solved by isolating a radical, squaring, isolating again, and squaring again.
- Quadratic roots are computed from Ax² + Bx + C = 0 (including stable handling when A ≈ 0).
- Every candidate is substituted back into the original radical equation to remove extraneous roots.
How to use this calculator
- Select the model matching your equation’s structure.
- Enter coefficients for each radicand and right side.
- Choose decimals and a tolerance for verification checks.
- Press Solve equation to see results above.
- Use CSV or PDF export buttons for sharing or saving.
Model selection and coefficient structure
This solver supports three common radical patterns: a single radical equated to a linear term, a two‑radical sum equated to a constant, and equality of two radicals. Each pattern uses coefficients a₁, b₁ (and a₂, b₂ when needed) to build radicands of the form a·x + b. For Model 1, the linear side uses c and d, allowing forms like √(3x+1)=2x−5. Clear mapping keeps inputs consistent across homework and exams.
Domain restrictions that prevent invalid roots
Real square‑roots require nonnegative radicands, so the calculator first derives inequalities such as a₁x+b₁≥0 and a₂x+b₂≥0. Model 1 adds cx+d≥0 because √(u)=v demands v≥0. These constraints often eliminate large parts of the number line. For example, √(x−4)=x−1 needs x≥4 and x≥1, so x≥4 before any algebra is trusted.
Squaring workflow and polynomial reduction
After domain setup, the solver squares to remove radicals and reduces the problem to linear or quadratic equations. Model 1 produces (cx+d)²=a₁x+b₁, which expands to a quadratic in x. Model 3 squares once and becomes a linear equation a₁x+b₁=a₂x+b₂. Model 2 isolates a radical, squares, isolates again, then squares a second time, yielding a quadratic that captures all candidates.
Verification stage and numerical tolerance
Because squaring can introduce extraneous solutions, every candidate is substituted back into the original radical equation. The tolerance field controls how close two floating values must be to count as equal, which is helpful when coefficients are decimals. A smaller tolerance is stricter; a larger tolerance is more forgiving for rounded inputs. The “Verified solutions” list is therefore the safe set to report.
Result reporting, steps, and exports
Results are shown above the form for quick iteration, while optional step output documents the algebraic pathway used. The example table illustrates typical outcomes: one solution, multiple solutions, or none. When you need a record for notes or lab documentation, export to CSV for spreadsheets or PDF for sharing. Keeping the model name, inputs, candidates, and verified roots together makes checking work faster and more reproducible. In practice, Model 1 needs one squaring, Model 3 needs one, and Model 2 needs two, so expect 2–5 candidate checks depending on the discriminant for each case.
FAQs
Why are some candidates rejected?
Squaring can create extraneous roots that satisfy the squared equation but fail the original radical equation. The solver substitutes each candidate back into the starting model and keeps only values that match within your chosen tolerance.
How should I choose the tolerance value?
Use a small tolerance for exact integers and fractions, such as 1e-9 to 1e-6. If you enter rounded decimals or measurements, use a slightly larger value, like 1e-6 to 1e-4, to avoid false rejections.
What happens if a coefficient is zero?
Zero slopes or intercepts are allowed. The solver automatically falls back to linear solving when the quadratic term disappears, and it treats constant radicands carefully. Always watch the domain notes, because a constant negative radicand has no real solutions.
Can it solve more complicated radical equations?
It targets three structured forms that cover many textbook problems. For more complex equations, rearrange algebraically to match one of the models, or reduce radicals stepwise. The verification step still helps confirm whether your transformation introduced extra roots.
Why does Model 1 require cx + d ≥ 0?
If √(u) equals a real number v, then v must be nonnegative because square-roots are defined as principal nonnegative values. Enforcing cx+d≥0 prevents accepting a candidate where the right side is negative.
How do CSV and PDF exports capture my work?
After you solve, the last result is stored for the session. The export includes the chosen model, all numeric inputs, candidate roots, verified solutions, and brief notes, making it easy to paste into reports or share with classmates.