Free · Algebra and real roots
Diamond Sum and Product: Real Number Pairs
Find a real pair from its sum and product, or calculate the top product and bottom sum from two side values.
- Formula & worked example
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Calculator inputs
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How to use this calculator
- Enter the known values in the units shown. Results update as you type.
- Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
- Use the worked example to check the method. Reset restores the starting fields.
Private by default
Inputs and results stay in this browser tab. Bookify does not upload or store the values you enter.
Formula and method
A sum of 5 and product of 6 give the pair 2 and 3. Put 6 at the top, 5 at the bottom and 2/3 on the sides.
S=a+b; P=ab; x²-Sx+P=0; roots=(S±sqrt(S²-4P))/2
Worked example
Enter these known values and leave the other values blank.
- Bottom: sum
- 5
- Top: product
- 6
- Side A
- 2
- Side B
- 3
Assumptions and limitations
- The four values are real numbers. A negative discriminant S²−4P has no real pair; zero gives a repeated pair. Complex roots are outside this tool.
- Solving from sum and product assigns the smaller root to side A and larger root to side B. When entering both side values directly, either order is valid.
- A side value and a nonzero product can solve the other side by division. A zero side with zero product alone is underdetermined; also provide the sum or the other side.
- The hidden discriminant helper retains its principal-positive sum inverse, which does not choose a negative sum by itself. Negative sums are supported by entering the sum directly or a determined side pair.
Common questions
Can one of the side numbers be negative?
Yes. Signed and fractional real numbers are accepted when the sum and product are consistent.
Why is there no real answer?
A negative S²−4P makes the square root imaginary. No pair of real side values has that sum and product.
References
The two real roots may be entered in either side order. Use symmetric product/division identities for consistency, with explicit zero-side sum cases. When solving from sum and product, obtain the smaller-magnitude root from product divided by the larger-magnitude root to avoid subtracting nearly equal numbers; use exact double-root and strict sign branches. The other 13 algebraic inverse variants remain unchanged. Reject strictly negative discriminants without relational tolerance, and show the fractional-number explanation when either side value is noninteger rather than only when both are.
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