Quadratic Equation Solver

What's the key thing to understand about The Formula?

x = (−b ± √(b²−4ac)) ÷ 2a. The ± gives two solutions (roots). When the discriminant (b²−4ac) is positive: two real roots. When zero: one repeated root. When negative: two complex roots.

What should I know about Example?

Worked example: solve x² − 5x + 6 = 0. Identify a=1, b=−5, c=6. Calculate discriminant: b² − 4ac = 25 − 24 = 1. Since discriminant > 0, there are two distinct real roots. Apply the quadratic formula: x = (−b ± √discriminant) / (2a) = (5 ± 1) / 2. Root 1: x = (5+1)/2 = 3. Root 2: x = (5−1)/2 = 2. Verify by substituting: 3² − 5(3) + 6 = 9 − 15 + 6 = 0 ✓ and 2² − 5(2) + 6 = 4 − 10 + 6 = 0 ✓. Both values satisfy the original equation.

What should I know about the Vertex and Parabola Shape?

Vertex (turning point): x-coordinate = −b/(2a). y-coordinate: substitute x back into the equation, y = c − b²/(4a). For 2x² − 3x − 2: x = 3/4 = 0.75, y = 2(0.75)² − 3(0.75) − 2 = −3.125, so the vertex is (0.75, −3.125). The parabola opens upward (a > 0 — minimum) or downward (a < 0 — maximum). The axis of symmetry is the vertical line x = −b/(2a).

What should I know about Applications?

Quadratic equations model projectile motion, area problems, profit maximisation, and countless physics problems. The roots represent where the parabola y = ax² + bx + c crosses the x-axis.

What's the key thing to understand about Real-World Quadratic Applications?

Quadratic equations appear wherever relationships involve squared variables. Projectile motion: height h = ut - 1/2 gt squared. Revenue maximisation: revenue is a quadratic function of price that peaks at the optimal price-quantity combination. Area problems: finding dimensions of a rectangle with given area and perimeter constraints. The discriminant (b squared minus 4ac) determines the number of solutions: positive gives two real solutions, zero gives one repeated solution, and negative gives no real solutions — only complex ones.

What should I know about the Discriminant and the Nature of the Roots?

The discriminant, calculated as b² − 4ac from the standard quadratic form ax² + bx + c = 0, reveals how many real solutions a quadratic equation has before you even solve it. If the discriminant is positive (b² − 4ac > 0), the equation has two distinct real roots, meaning the parabola crosses the x-axis at two separate points. If the discriminant equals zero (b² − 4ac = 0), the equation has exactly one repeated real root, meaning the parabola just touches the x-axis at a single point (its vertex sits exactly on the x-axis). If the discriminant is negative (b² − 4ac < 0), the equation has no real roots — only two complex conjugate roots involving the imaginary unit i — meaning the parabola never crosses the x-axis at all. For example, x² − 4x + 4 = 0 has discriminant (−4)² − 4(1)(4) = 16 − 16 = 0, confirming a single repeated root at x = 2, which matches the fact that x² − 4x + 4 factors perfectly as (x−2)².

What's the difference between Factoring and the Quadratic Formula?

Factoring and the quadratic formula both solve ax² + bx + c = 0, but they differ in reliability and ease. Factoring rewrites the equation as a product of two binomials, such as x² − 5x + 6 = 0 becoming (x−2)(x−3) = 0, immediately giving roots x = 2 and x = 3 — this method is fast when the roots happen to be simple integers or fractions, but finding the right factors by inspection becomes difficult or impossible for equations with irrational or complex roots. The quadratic formula, x = (−b ± √(b²−4ac)) / 2a, always works regardless of whether the roots are whole numbers, fractions, irrational numbers, or complex numbers, making it the universally reliable method — the trade-off is that it requires more arithmetic, particularly when the discriminant isn't a perfect square. In practice, it's often fastest to glance at an equation for obvious integer factors first, and fall back on the quadratic formula whenever factoring isn't immediately apparent, which is exactly why this calculator applies the formula directly rather than searching for factors.

What do I need to know about Sum and Product of Roots (Vieta's Formulas)?

For a quadratic ax² + bx + c = 0 with roots p and q, Vieta's Formulas state that the sum of the roots equals -b/a, and the product of the roots equals c/a — without needing to solve for the roots individually first. This gives a fast way to check a solution: after solving a quadratic, adding the two roots should equal -b/a, and multiplying them should equal c/a; if either check fails, there's an arithmetic error somewhere. It also works in reverse — given two desired roots, you can construct the quadratic equation that produces them: a quadratic with roots 3 and -5 is x² - (3 + (-5))x + (3×(-5)) = x² + 2x - 15. This relationship is a cornerstone of more advanced algebra and is frequently tested directly in exams as "find the value of p² + q²" or similar, using sum and product rather than the individual root values.

Worked Example: Sum and Product of Roots Without Solving

For the quadratic 2x² - 7x + 3 = 0, Vieta's formulas give the sum and product of the roots directly from the coefficients, without solving for the roots individually: sum = -b/a = -(-7)/2 = 3.5, and product = c/a = 3/2 = 1.5. Checking against the actual roots (found via the quadratic formula: x = [7 ± √(49-24)]/4 = [7 ± 5]/4, giving x = 3 and x = 0.5): sum = 3 + 0.5 = 3.5 ✓, product = 3 × 0.5 = 1.5 ✓. This trick is especially useful for quickly verifying a quadratic formula answer, or for constructing a quadratic equation when only the two roots are known — if roots are r₁ and r₂, the quadratic is simply x² - (r₁+r₂)x + r₁r₂ = 0 (scaled by any constant a).

Worked Example: Quadratic Model of Projectile Height

A ball's height in metres is modelled by h(t) = -5t² + 20t + 1, where t is time in seconds (using g ≈ 10 m/s² for simplicity). The maximum height occurs at the vertex, t = -b/(2a) = -20/(2×-5) = -20/-10 = 2 seconds. The maximum height itself is h(2) = -5(2)² + 20(2) + 1 = -20 + 40 + 1 = 21 m. This vertex calculation is identical in method to finding the vertex of any parabola y = ax² + bx + c, just with physical labels attached: a relates to gravitational deceleration, b to initial velocity, and c to initial height. To find when the ball lands (h = 0), the quadratic formula gives t = [-20 ± √(400+20)]/(-10) = [-20 ± √420]/(-10) ≈ [-20 ± 20.49]/(-10), taking the positive root: t ≈ 0.049 s before launch is discarded, giving a landing time of about t ≈ 4.05 s — roughly double the time to reach maximum height, which is expected since a projectile under constant gravity takes the same time to fall back down as it took to climb, adjusted slightly here by the nonzero starting height.

Worked Example: A Quadratic With Complex Roots

Solve x² + 2x + 5 = 0. The discriminant is b² - 4ac = 2² - 4(1)(5) = 4 - 20 = -16, which is negative, meaning there are no real solutions — the parabola never crosses the x-axis. The roots are complex: x = [-2 ± √(-16)]/2 = [-2 ± 4i]/2 = -1 ± 2i. So the two complex roots are x = -1 + 2i and x = -1 - 2i. Notice they are complex conjugates of each other (same real part, opposite sign imaginary part) — this always happens for a quadratic with real coefficients and a negative discriminant, since the ± in the quadratic formula only ever flips the sign of the imaginary part, never introduces an entirely different real part.

Checking a Solution by Substitution

After solving a quadratic, always verify by substituting a root back into the original equation. For 2x² - 7x + 3 = 0 with root x = 3: 2(3)² - 7(3) + 3 = 2(9) - 21 + 3 = 18 - 21 + 3 = 0 ✓. This simple check catches arithmetic slips in the quadratic formula (a common source of error is mis-squaring a negative b, or dropping a negative sign under the square root) before they propagate into a wrong final answer, and it takes only a few seconds compared to the time spent solving the equation in the first place.