Quadratic equations

A quadratic equation is one in which the unknown appears in the second power. Once tidied up, every one of them can be written in the general form:

$$ax^2 + bx + c = 0 \quad (a \neq 0)$$

The condition $a \neq 0$ is not decoration — it is what tells a quadratic equation from a linear one. The numbers $a$, $b$ and $c$ are called coefficients; $b$ or $c$ may be zero, but $a$ never is.

The discriminant

Before computing anything else, we compute one number:

$$\Delta = b^2 - 4ac$$

That number sits under the square root in the formulas for the solutions, so its sign settles everything before the search for the roots even begins:

• $\Delta > 0$ — two distinct roots,
• $\Delta = 0$ — one (double) root,
• $\Delta < 0$ — no real roots.

The formulas for the roots

When $\Delta > 0$, both solutions are given by:

$$x_1 = \frac{-b - \sqrt{\Delta}}{2a}, \qquad x_2 = \frac{-b + \sqrt{\Delta}}{2a}$$

When $\Delta = 0$, both formulas give the same number, so they are written more briefly:

$$x_0 = \frac{-b}{2a}$$

The commonest slip lives in the signs: the formula has $-b$, so a $b = -2$ turns into a $+2$ in the numerator.

Incomplete equations

When the middle or the constant term is missing, the discriminant of course still works — but it is a waste of time, because the shortcut is shorter:

$$x^2 - 9 = 0 \iff x^2 = 9 \iff x = 3 \ \text{ or } \ x = -3$$ $$x^2 - 5x = 0 \iff x(x - 5) = 0 \iff x = 0 \ \text{ or } \ x = 5$$

The first case is taking the square root of both sides — with both signs, because squaring loses the sign. The second is factoring $x$ out and using the fact that a product is zero only when one of its factors is. That is the same rule the factored form rests on:

$$ax^2 + bx + c = a(x - x_1)(x - x_2)$$

The parabola and its zeros

The graph of $y = ax^2 + bx + c$ is a parabola. So the equation $ax^2 + bx + c = 0$ asks where that parabola has height zero — that is, where it crosses the $x$-axis:

The three cases of the discriminant are now plain to see: for $\Delta > 0$ the parabola crosses the axis twice, for $\Delta = 0$ it touches it once, and for $\Delta < 0$ it misses it entirely — lying wholly above or wholly below.

The parabola also has a vertex, an axis of symmetry and a canonical form, but those are properties of the quadratic function — we come back to them in the branch on functions. Here only one thing about it matters: where it meets the $x$-axis.

Vieta formulas

The sum and the product of the roots can be read off the coefficients alone, without computing the discriminant:

$$x_1 + x_2 = -\frac{b}{a}, \qquad x_1 \cdot x_2 = \frac{c}{a}$$

For $x^2 - 2x - 3 = 0$ that gives a sum of $2$ and a product of $-3$ — matching the pair $-1$ and $3$. It is the cheapest check of a result there is, and with whole-number coefficients it often lets you guess the roots: look for two numbers with the given sum and the given product.

In the exercises below, enter both roots separated by a comma, in ascending order — for instance $x_1 = -1, x_2 = 3$ (a bare $-1, 3$ is accepted too). When $\Delta = 0$ and there is a single root, enter a single number.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

x² − 11x + 30 = 0ResultCheck

Common mistakes

• Computing the discriminant before tidying the equation up — move everything to one side first, then read off $a$, $b$, $c$.
• A lost sign in $-b$ — for $b = -2$ the numerator starts with $+2$.
• Dividing only the first term by $2a$ — the fraction bar covers the whole numerator.
• Keeping one sign when taking a root — $x^2 = 9$ gives $x = 3$ or $x = -3$.
• Dividing the equation by $x$ — in $x^2 = 5x$ that loses the solution $x = 0$; factor $x$ out instead.
• Reading a negative discriminant as an arithmetic slip — it is a correct answer: no real roots.