Solving quadratic and simultaneous equations
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The quadratic formula and how the discriminant tells you whether roots are real, repeated or complex; when a system of linear equations has no solution or infinitely many; and how to use the equation solver.
The roots of ax²+bx+c = 0 are x = (−b ± √(b²−4ac)) / 2a. If the discriminant D = b²−4ac is positive there are two real roots, if it is 0 one repeated root, and if negative two complex roots. Enter a, b and c on the equation screen to get exact roots like 1+√2.
Quadratic examples
| Equation | Discriminant D | Roots |
|---|---|---|
| x² − 2x − 1 = 0 | 8 (positive) | x = 1 − √2, 1 + √2 (about −0.4142135624, 2.414213562) |
| x² − 6x + 9 = 0 | 0 | x = 3 (repeated) |
| x² + 2x + 5 = 0 | −16 (negative) | x = −1 − 2i, −1 + 2i |
For the first one, the formula gives x = (2 ± √8)/2 = (2 ± 2√2)/2 = 1 ± √2. The calculator simplifies √8 to 2√2 for you.
The graph of y = x² − 2x − 1 is lowest at x = 1, where y = −2 (vertex (1, −2)). The equation screen shows the vertex along with the roots.
Solve x²−2x−1=0 on the equation screen ▸
Cubics and quartics
From degree three on, solving by hand gets hard. The calculator first looks for whole-number and fraction roots to lower the degree, then solves what is left with the quadratic formula or numerically. For example, x³ − 6x² + 11x − 6 = 0 has roots 1, 2 and 3.
Simultaneous linear equations
Solving 2x + 3y = 12 and x − y = 1 gives x = 3, y = 2. From the second equation x = y + 1; substituting, 2(y + 1) + 3y = 12, so 5y = 10 and y = 2.
A system does not always have exactly one solution:
- No solution: x + y = 1 and 2x + 2y = 3. The left sides are proportional but the right sides are not (parallel lines).
- Infinitely many: x + y = 1 and 2x + 2y = 2. The two equations are really the same line.
The equation screen reports these cases as “no solution” and “infinitely many solutions”. It handles up to four unknowns.
Tips
- Enter coefficients from the highest power down, with 0 for missing terms: x² − 4 = 0 is a = 1, b = 0, c = −4.
- Boxes accept expressions such as 1/3, √2 or 2π.
- To check a root, substitute it back into the original equation and see that you get 0.
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