Roots, Discriminant & Vertex
Last reviewed: May 2026
x = (−b ± √(b²−4ac)) / 2a solves any equation ax² + bx + c = 0. It finds both roots, reveals the discriminant (number of real solutions), and locates the vertex. Whether the equation factors cleanly or not, this formula always works.1
| Discriminant | Real Roots | Graph | Example |
|---|---|---|---|
| Positive | 2 distinct | Crosses x-axis twice | x²−5x+6=0 → x=2,3 |
| Zero | 1 repeated | Touches x-axis | x²−6x+9=0 → x=3 |
| Negative | 0 real (2 complex) | No x-axis crossing | x²+x+1=0 |
The quadratic formula solves any equation of the form ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) / 2a. It works when factoring fails and always produces the correct answer. The "±" indicates two solutions — a parabola crosses the x-axis at most twice. The formula is derived by completing the square on the general quadratic equation, a technique documented in Babylonian mathematics over 4,000 years ago.
The expression under the square root, b² − 4ac, is called the discriminant and reveals the nature of solutions before you calculate them. b² − 4ac > 0: Two distinct real solutions — the parabola crosses the x-axis twice. b² − 4ac = 0: One repeated real solution — the parabola touches the x-axis at its vertex. b² − 4ac < 0: No real solutions (two complex/imaginary solutions) — the parabola doesn't reach the x-axis. Checking the discriminant first saves time — if it's negative, there's no need to compute further for real-number applications.
Example 1: Solve 2x² + 5x − 3 = 0. Here a=2, b=5, c=−3. Discriminant = 25 − 4(2)(−3) = 25 + 24 = 49. √49 = 7. x = (−5 ± 7) / 4. Solutions: x = (−5+7)/4 = 0.5 and x = (−5−7)/4 = −3.
Example 2: Solve x² − 6x + 9 = 0. Discriminant = 36 − 36 = 0. One solution: x = 6/2 = 3. The parabola just touches the x-axis at x=3.
Example 3: Solve x² + 2x + 5 = 0. Discriminant = 4 − 20 = −16 < 0. No real solutions — the parabola y = x²+2x+5 has its vertex above the x-axis and never crosses it.
Projectile motion: The height of a thrown ball follows h = −16t² + v₀t + h₀. Finding when it hits the ground (h=0) requires the quadratic formula. Area optimization: "A farmer has 200 feet of fencing to enclose a rectangular area. What dimensions maximize the area?" leads to a quadratic. Revenue modeling: If price affects demand linearly, revenue (price × quantity) is quadratic, and maximizing revenue requires finding the vertex. Engineering: Beam deflection, electrical circuit analysis (RLC circuits), and lens equations all produce quadratic relationships.
Factoring: If the quadratic factors neatly (x²−5x+6 = (x−2)(x−3)), this is faster. Works well when coefficients are small integers. Completing the square: Rewrite ax²+bx+c as a(x−h)²+k to find the vertex form. More steps but reveals the parabola's shape. Graphing: Plot y = ax²+bx+c and read x-intercepts. Good for estimation and visualization. The quadratic formula is the universal method — it works when these alternatives don't, and it's guaranteed to produce all solutions.
The quadratic formula x = (-b ± √(b²-4ac)) ÷ 2a solves any equation of the form ax² + bx + c = 0. It comes from completing the square on the general quadratic. Starting with ax² + bx + c = 0: divide by a to get x² + (b/a)x + c/a = 0. Move the constant: x² + (b/a)x = -c/a. Add (b/2a)² to both sides to complete the square: (x + b/2a)² = b²/4a² - c/a = (b² - 4ac)/4a². Take the square root: x + b/2a = ±√(b²-4ac)/2a. Subtract b/2a: x = (-b ± √(b²-4ac))/2a. Understanding this derivation matters because it reveals that the formula isn't magical — it's a systematic application of algebra that students can reproduce from scratch. If you forget the formula on an exam, you can always complete the square on the specific equation instead.
The expression under the square root, b² - 4ac (called the discriminant), tells you everything about the nature of solutions before you solve. When b² - 4ac > 0, the equation has two distinct real solutions — the parabola crosses the x-axis at two points. When b² - 4ac = 0, there's exactly one repeated real solution — the parabola just touches the x-axis at its vertex. When b² - 4ac < 0, there are no real solutions (but two complex solutions involving imaginary numbers) — the parabola never reaches the x-axis. For example: x² - 5x + 6 = 0 has discriminant 25 - 24 = 1 > 0 (two solutions: x = 2 and x = 3). x² - 4x + 4 = 0 has discriminant 16 - 16 = 0 (one solution: x = 2). x² + x + 1 = 0 has discriminant 1 - 4 = -3 < 0 (no real solutions). Engineers and physicists check the discriminant first to determine whether a physical system has real solutions before investing time in computation.
Quadratic equations model any situation involving area, projectile motion, or optimization. A farmer with 200 feet of fencing wants to enclose a rectangular area along a river (one side needs no fence). If the width is x, the length is 200 - 2x, and area = x(200-2x) = -2x² + 200x. Maximum area occurs at the vertex: x = -200/(2×(-2)) = 50 feet, giving a 50��100 foot rectangle with 5,000 sq ft. Projectile height follows h(t) = -16t² + v₀t + h₀ (in feet, with t in seconds). A ball thrown upward at 64 ft/s from 6 feet high: h(t) = -16t² + 64t + 6. It hits the ground when h = 0: using the formula, t = (-64 ± √(4096 + 384))/(-32) = (-64 ± 66.94)/(-32). The positive root gives t ≈ 4.09 seconds. Revenue optimization is another common application: if a company charges p dollars per unit and sells (1000 - 20p) units, revenue = p(1000-20p) = -20p² + 1000p, maximized at p = 1000/40 = $25 with revenue of $12,500.
The quadratic formula always works, but it's not always the most efficient approach. Factoring is faster when the equation factors neatly: x² - 7x + 12 = (x-3)(x-4), giving x = 3 and x = 4 without formula computation. For equations where a = 1 and b and c are small integers, factoring takes seconds versus 30+ seconds with the formula. Completing the square is preferred when you need the vertex form y = a(x-h)² + k, which directly reveals the vertex (h,k) of the parabola — essential for graphing and optimization problems. Graphical methods using a calculator or graphing software show solutions as x-intercepts and provide visual context about the parabola's shape and direction. The quadratic formula is the universal fallback: use it whenever factoring isn't obvious, when coefficients are large or non-integer, or when you need exact irrational answers (like x = (3 + √17)/4) rather than decimal approximations.
Several errors recur frequently when students and professionals work with quadratic equations. The most damaging: dropping the negative sign on the -b term. In x² + 6x + 5 = 0, b = 6, so -b = -6 — but writing +6 instead produces wrong roots. Another common error: dividing only part of the numerator by 2a. The fraction bar in (-b ± √(b²-4ac))/2a applies to the entire numerator, not just the square root. Students sometimes compute -b/2a + √(b²-4ac), forgetting to divide the radical part by 2a as well. Sign errors inside the discriminant trip up many: b²-4ac for equation 3x² - 5x + 2 = 0 is (-5)² - 4(3)(2) = 25 - 24 = 1, not (-5)² - 4(3)(2) = -25 - 24. Remember that (-5)² = 25, not -25. Finally, forgetting the ± symbol produces only one solution instead of two — always compute both the + and - cases unless the discriminant is zero.
Quadratic equations appear throughout financial modeling. The break-even point for a business with fixed costs F, variable cost per unit v, and revenue per unit modeled as p = a - bq (price decreases as quantity increases) requires solving the quadratic revenue = cost equation: q(a-bq) = F + vq, or -bq² + (a-v)q - F = 0. Bond duration — a measure of interest rate sensitivity — involves quadratic terms in its convexity adjustment. Option pricing models (Black-Scholes) contain quadratic terms in the exponent of the normal distribution. Even simple loan calculations can generate quadratics: if you want to find the interest rate that makes monthly payments of $500 pay off a $20,000 loan in exactly 48 months, the resulting equation after simplification contains a quadratic relationship between rate and payment. While financial calculators solve these iteratively, understanding the underlying quadratic structure helps interpret why some financial problems have two solutions (two break-even points) and others have none (a business model that can never break even).
The quadratic formula solves degree-2 polynomials. Cubic (degree 3) and quartic (degree 4) equations also have general formulas, discovered in the 16th century by Cardano and Ferrari, but they are far more complex and rarely used by hand. The Abel-Ruffini theorem proves that no general formula exists for degree 5 and higher — a landmark result in abstract algebra. In practice, numerical methods (Newton-Raphson iteration, the bisection method) solve higher-degree equations computationally. However, many real-world problems can be reduced to quadratics through substitution: the equation x⁴ - 5x² + 4 = 0 becomes u² - 5u + 4 = 0 (where u = x²), yielding u = 1 or u = 4, so x = ±1 or x = ±2. Recognizing these "quadratic in disguise" patterns extends the quadratic formula's reach well beyond standard ax² + bx + c problems.
→ Check the discriminant first. Saves time — negative means no real solutions.
→ Verify by plugging back in. Substitute roots into the original equation to confirm.
→ Watch signs. The most common error is sign mistakes in b or c.
→ Use for optimization. The vertex gives max/min — useful for real-world optimization problems.
See also: Scientific Calculator · Matrix · Statistics · Percentage