"Solving" a quadratic equation means finding every -value that makes it true — geometrically, these are exactly the -intercepts of the parabola . Quadratic equations appear whenever a question asks "when" or "where" about a quadratic model: when does a ball hit the ground, what price gives zero profit, what width gives an area of , or where does a straight path cross a curved one. There are several tools, and a strong student picks the fastest one for the equation in front of them. Sometimes factoring is fastest; sometimes the numbers are too messy and you need a formula that works every time; and sometimes you only need to know how many solutions exist, which the discriminant tells you without solving at all.
Method 1: Factoring (and square roots)
If factors nicely, set each factor equal to zero using the zero product property: if , then or . The property works only for zero, because zero is the one number you cannot get by multiplying two non-zero numbers. That is why the equation must be set equal to zero first. For , writing tells you nothing (many pairs multiply to ). Instead rearrange to , factor as , and read off or . Check by substituting: ✓ and ✓.
Two special shapes are faster still. If there is no -term, isolate the square and take square roots, remembering both signs: gives , so ; and gives , so . If there is no constant term, factor out : becomes , so or . Never divide both sides by here, because that throws away the solution .
Method 2: The quadratic formula
The formula is not magic: it is what you get by completing the square on the general equation . Divide by , move across, add to both sides, and you reach . Taking the square root of both sides (with ) and subtracting gives the formula. To use it, identify , , from standard form, substitute with brackets around negative values, and simplify. Give exact answers when asked, otherwise round at the end. For : , which is about or .
The discriminant
The expression under the square root, , is called the discriminant. You don't need to finish solving to know how many real roots exist, because the square root decides everything. If , is a positive number, so and give two different roots. If , the adds and subtracts zero, so both roots collapse into one: , which is the vertex sitting on the -axis. If , there is no real square root of a negative number, so there are no real roots and the parabola never reaches the -axis. One more clue: if is a perfect square (like or ), the roots are rational and the quadratic factors over the integers, so factoring would have worked.
| Discriminant | Number of real roots | What the graph looks like |
|---|---|---|
| 2 distinct real roots | crosses the x-axis twice | |
| 1 real root (repeated) | touches the x-axis once (vertex on the axis) | |
| 0 real roots | never touches the x-axis |
Where a line meets a parabola
To find where the line meets the parabola , look for points that satisfy both equations: set the right sides equal, move everything to one side and solve the resulting quadratic. For and : becomes , so and or . Substitute into the simpler (linear) equation for the -values: the points are and . The discriminant of that combined quadratic counts the intersections: means the line cuts the parabola twice (a secant), means it touches once (a tangent line), and means they never meet. For and , the combined equation is with , so the line is tangent exactly when .
- Fastest when the coefficients are small whole numbers
- Works only when the roots are rational (D is a perfect square)
- Gives exact answers with no radicals to simplify
- Example:
- Works on every quadratic, including ones that do not factor
- Slower, and sign errors are the main risk
- Exact answers may contain radicals, like
- Its part doubles as the root-count test
In applied problems, always check that each root makes sense. A time, length or price cannot be negative, so a root like s is rejected even though it is correct algebra. State both roots, then say which one answers the question and why.