Number & Algebra
Quadratic equations
The discriminant, nature of roots, and the graph of y = ax² + bx + c.
A quadratic equation in one unknown has the form \(ax^2 + bx + c = 0\), where \(a \neq 0\). You can solve these by factorisation, the quadratic formula, or by reading information from the graph of \(y = ax^2 + bx + c\).
The discriminant
The discriminant \(\Delta\) (delta) is defined below. It tells you how many real roots the equation has without fully solving it.
Definition
\[ \Delta = b^2 - 4ac \]
Nature of roots
| Discriminant | Nature of roots | Graph of \(y = ax^2 + bx + c\) |
|---|---|---|
| \(\Delta > 0\) | Two distinct real roots | Parabola cuts the x-axis twice |
| \(\Delta = 0\) | One repeated real root (equal roots) | Vertex touches the x-axis |
| \(\Delta < 0\) | No real roots | Parabola lies entirely above or below the x-axis |
When \(\Delta \geq 0\), the quadratic formula gives the roots: \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \]
Try it yourself
Use the explorer below to change \(a\), \(b\), and \(c\). Watch how \(\Delta\) and the parabola change, and read off the nature of the roots.
History
People have worked on problems that lead to quadratic equations for thousands of years. Babylonian clay tablets from around 1800 BCE record area and length puzzles — for example, finding the sides of a rectangle when you know its area and perimeter. Those situations produce equations in one unknown whose highest power is two.
Early methods were often geometric: you imagine completing a rectangle to make a square, which is the same idea behind completing the square today. In the 9th century, al-Khwarizmi wrote systematic rules for many types of quadratic problems, sorted by which coefficients were present. He did not use modern symbols like \(x^2\), but the logic matches what we still teach.
During the Renaissance, mathematicians such as Cardano and Tartaglia pushed further into solving cubics and quartics. Along the way they met equations whose discriminants were negative — no real roots, yet the algebra still demanded answers. That tension helped open the door to complex numbers, a topic you will meet in a later chapter.
Derivation
The quadratic formula comes from rearranging \(ax^2 + bx + c = 0\) and completing the square. Below is a guided derivation; routine algebra between steps is left for you to check.
Start with \(a \neq 0\) and divide every term by \(a\): \[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0. \] Move the constant to the right: \[ x^2 + \frac{b}{a}x = -\frac{c}{a}. \]
The expression \(x^2 + \frac{b}{a}x\) is almost a perfect square. Add and subtract \(\left(\frac{b}{2a}\right)^2\) on the left (equivalently, add that square to both sides): \[ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2. \] The left side factors as a square: \[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}. \]
Take square roots (when the right-hand side is non-negative you get real solutions): \[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}. \] Solve for \(x\):
Quadratic formula
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The quantity under the square root is exactly the discriminant: \(\Delta = b^2 - 4ac\). So the formula is \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\) whenever \(\Delta \geq 0\). If \(\Delta > 0\) you get two distinct real roots; if \(\Delta = 0\) the \(\pm\) gives the same root twice; if \(\Delta < 0\) there is no real square root and the equation has no real solutions — matching the nature-of-roots table above.
Checkpoints
Pause and think before moving on. There are no answers on this page — discuss with a classmate or write your reasoning in a notebook.
Checkpoint
In the explorer, keep \(a\) and \(b\) fixed and slowly change \(c\) so that \(\Delta\) goes from positive to zero. What happens to the two x-intercepts on the graph?
Checkpoint
Can \(ax^2 + bx + c = 0\) with \(a \neq 0\) have three distinct real solutions? Why or why not?
Checkpoint
Suppose \(a > 0\) and \(\Delta < 0\). Where does the entire parabola \(y = ax^2 + bx + c\) sit relative to the x-axis?
Checkpoint
A ball is thrown straight upward from ground level. Without using numbers, describe a situation where \(\Delta > 0\) for height versus time, and another where \(\Delta = 0\).
Checkpoint
After completing the square you had \(\left(x + \frac{b}{2a}\right)^2 = \frac{\Delta}{4a^2}\). Why must the sign of \(\Delta\) match whether the parabola crosses the x-axis?
Applications
Projectile motion
Near Earth’s surface, height \(h\) (in metres) of an object thrown upward often follows \[ h = -\tfrac{1}{2}gt^2 + v_0 t + h_0, \] where \(t\) is time in seconds, \(g \approx 9.8\) is gravitational acceleration, \(v_0\) is initial upward speed, and \(h_0\) is starting height. This is a quadratic in \(t\). Setting \(h = 0\) asks when the object is at ground level; the solutions are the landing times. The discriminant tells you whether there are two crossing times (\(\Delta > 0\)), one (\(\Delta = 0\)), or none (\(\Delta < 0\)) — for example if the model never reaches the ground in the time you care about.
Profit and pricing
A shop might find that when it charges \(p\) dollars per item, revenue is \(p\) times quantity sold, while cost has a fixed part plus a cost per item. Profit (revenue minus cost) is often a quadratic in \(p\) that opens downward — there is a best price at the vertex. For example, with profit \(\Pi\) in dollars, \[ \Pi(p) = -2p^2 + 40p - 150. \] Setting \(\Pi = 0\) asks for break-even prices; \(\Delta\) tells you whether two break-even prices exist, one, or none.
Area with fixed perimeter
A rectangle with perimeter \(P\) can be described by width \(x\) and length \(\frac{P}{2} - x\). Its area is \[ A(x) = x\left(\frac{P}{2} - x\right) = -\!x^2 + \frac{P}{2}x, \] a downward-opening parabola. The largest area occurs at the vertex, not at the extremes \(x = 0\) or \(x = \frac{P}{2}\) (which give zero area).
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