Solve quadratic equations in the form ax² + bx + c = 0. Get step-by-step solutions, discriminant analysis, and vertex calculations.
A quadratic equation is a second-degree polynomial equation in the form ax² + bx + c = 0, where:
Examples of quadratic equations:
Quadratic equations appear throughout mathematics, science, and engineering:
The quadratic formula solves any quadratic equation. It's derived by completing the square on the general form ax² + bx + c = 0.
-b: Negative of the b coefficient
±: Plus or minus—gives two solutions (x₁ with +, x₂ with -)
b² - 4ac: The discriminant (Δ), determines number and type of solutions
√(b² - 4ac): Square root of discriminant
2a: Twice the a coefficient (denominator for entire expression)
Example: Solve x² - 5x + 6 = 0
The discriminant Δ = b² - 4ac determines the number and type of solutions without actually solving the equation.
Two distinct real solutions
Example: x² - 5x + 6 = 0
Δ = (-5)² - 4(1)(6) = 25 - 24 = 1 > 0
Solutions: x = 2 and x = 3 (two different real numbers)
Graph: Parabola crosses x-axis at two points
One repeated real solution (double root)
Example: x² - 6x + 9 = 0
Δ = (-6)² - 4(1)(9) = 36 - 36 = 0
Solution: x = 3 (only one solution, with multiplicity 2)
Graph: Parabola touches x-axis at exactly one point (vertex on x-axis)
No real solutions (two complex/imaginary solutions)
Example: x² + 2x + 5 = 0
Δ = 2² - 4(1)(5) = 4 - 20 = -16 < 0
No real solutions (solutions involve imaginary numbers: x = -1 ± 2i)
Graph: Parabola doesn't cross x-axis
If discriminant is a perfect square (1, 4, 9, 16, 25...), solutions are rational numbers (fractions or integers). Otherwise, solutions involve irrational numbers (square roots).
Δ = 1: Solutions involve ±√1 = ±1 (rational)
Δ = 5: Solutions involve ±√5 ≈ ±2.236... (irrational)
If quadratic factors nicely, factoring is fastest method.
Example: x² - 5x + 6 = 0
Limitation: Only works when equation factors with integers. Many quadratics don't factor nicely (e.g., x² + x - 1 = 0).
Rewrites equation as perfect square, then solves. Method used to derive quadratic formula.
Example: x² + 6x + 5 = 0
Use case: Good for understanding vertex form and derivations. Less practical than quadratic formula for computation.
Graph y = ax² + bx + c and find x-intercepts (where graph crosses y = 0).
Pros: Visual understanding, approximate solutions quickly
Cons: Less precise, requires graphing technology, doesn't show exact irrational solutions
Try factoring first if coefficients are small integers—fastest if it works.
Use quadratic formula if factoring doesn't work or coefficients are complex—always works, gives exact answers.
Complete the square for deriving vertex form or theoretical work.
Graph for visual understanding or approximate solutions.
The vertex is the highest or lowest point on the parabola.
Formula: Vertex = (-b/(2a), f(-b/(2a)))
x-coordinate: x = -b/(2a)
y-coordinate: Substitute x into equation to find y
Example: y = x² - 4x + 3
Vertical line through vertex: x = -b/(2a)
Parabola is symmetric—mirror image across this line.
a > 0: Parabola opens upward (∪ shape). Vertex is minimum point.
a < 0: Parabola opens downward (∩ shape). Vertex is maximum point.
|a| larger: Narrower parabola (steep sides)
|a| smaller: Wider parabola (gentle curve)
Where parabola crosses y-axis (when x = 0).
Formula: y-intercept = c
Simply the constant term in ax² + bx + c.
Where parabola crosses x-axis (when y = 0).
These are the solutions to ax² + bx + c = 0—what quadratic formula finds.
Ball thrown upward: h(t) = -16t² + v₀t + h₀
h = height, t = time, v₀ = initial velocity, h₀ = initial height
Example: Ball thrown at 64 ft/s from 6 ft height. When does it hit ground?
-16t² + 64t + 6 = 0
Using quadratic formula: t ≈ 4.09 seconds
Negative solution (t ≈ -0.09) is discarded—time can't be negative.
Rectangular garden 40 ft perimeter. What dimensions maximize area?
If width = x, length = 20 - x (since perimeter = 2w + 2l = 40)
Area = x(20 - x) = 20x - x² = -x² + 20x
Vertex gives maximum: x = -20/(2×-1) = 10 ft
Maximum area when width = length = 10 ft (square maximizes area)
Profit function: P(x) = -2x² + 80x - 200
x = number of units, P = profit in dollars
Break-even points: -2x² + 80x - 200 = 0
Solutions: x = 5 and x = 35 (break even at 5 units and 35 units)
Maximum profit: Vertex at x = -80/(2×-2) = 20 units
P(20) = -2(400) + 80(20) - 200 = $600 maximum profit
Bridge arch, satellite dish shape, headlight reflectors—all parabolic. Quadratic equations model these curves for engineering calculations.
❌ Wrong: For b = -5, using -b = -5 in formula
✓ Correct: -b = -(-5) = 5
Carefully handle negative signs, especially with -b.
❌ Wrong: b² - 4ac = -5² - 4(1)(6) = -25 - 24
✓ Correct: b² - 4ac = (-5)² - 4(1)(6) = 25 - 24
Square b first before applying negative. (-5)² = 25, not -25.
❌ Wrong: x = (-b + √discriminant) / 2a (only one solution)
✓ Correct: x = (-b ± √discriminant) / 2a (two solutions)
Always calculate both + and - to get both roots.
❌ Wrong: x = -b / 2a ± √discriminant
✓ Correct: x = (-b ± √discriminant) / 2a
Entire numerator divides by 2a, not just -b.
❌ Leaving answer as: x = (6 ± √8) / 2
✓ Better: √8 = 2√2, so x = (6 ± 2√2) / 2 = 3 ± √2
Simplify radicals and reduce fractions for cleaner final answer.
If a = 0, equation is linear (bx + c = 0), not quadratic. Quadratic formula doesn't apply. Solve as x = -c/b instead.
No real solutions exist—solutions are complex numbers involving √-1 = i. For basic algebra and real-world problems, "no real solutions" is the answer. In advanced math, calculate complex solutions: x = (-b ± i√|discriminant|) / 2a.
No. Quadratic equations have maximum two solutions (or one repeated solution). Three solutions would require cubic equation (x³). This is fundamental algebra theorem.
Try factoring first for simple integer coefficients—faster if it works. If can't factor in 30 seconds, use quadratic formula—always works and gives exact answer. Quadratic formula is more reliable for complex coefficients, decimals, or fractions.
The ± in quadratic formula means we use both positive and negative square roots—that's why we get two solutions. In other contexts (like Pythagorean theorem for side length), we use only positive because negative length doesn't make physical sense.
Check both solutions against problem context. Negative time, negative length, or values outside stated domain should be discarded. Example: If solving for time and get t = 5 or t = -2, discard t = -2 (time can't be negative). Use t = 5.
All same thing, different terminology: Solutions to ax² + bx + c = 0. Roots of the equation. Zeros of function f(x) = ax² + bx + c. X-intercepts of parabola y = ax² + bx + c. Different names for where parabola crosses x-axis.