Quadratic Equations

Throw a ball upward and it traces a curved arc before falling back; the jet of a fountain and the dive of a roller coaster are all relatives of that curve. It is called a parabola, and the director behind the scenes is an equation carrying $x^2$ . And $x^2$ is just $x \cdot x$ — the multiplication shortcut from Exponents, taking the lead role today.

What makes an equation quadratic

Look at the highest power of x in the expression:

$2x + 1 = 0$ : highest power is 1 — straight-line business, governed by Equation of a Straight Line;
$x^2 - 3x - 10 = 0$ : highest power is $x^2$ — this is a quadratic equation.

In general form it reads $ax^2 + bx + c = 0$ (with $a \ne 0$ ). The moment $a = 0$ , the $x^2$ term vanishes and the equation falls back to a line.

Make a table, and the parabola appears

Build a table for $y = x^2$ :

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$9$	$4$	$1$	$0$	$1$	$4$	$9$

Two observations:

y is never negative — squaring never goes into debt;
x at $-3$ and $3$ both give $y = 9$ : the graph is mirror-symmetric and bends into a bowl. That bowl is the parabola.

Solving it: factor first, then conquer each

Solve $x^2 - 3x - 10 = 0$ . The left side factors into two brackets — find two numbers whose product is $-10$ and whose sum is $-3$ : they are $-5$ and $2$ . So

$(x - 5)(x + 2) = 0$

The key step: if two numbers multiply to 0, at least one of them is 0. So either

$x - 5 = 0$

$x + 2 = 0$

giving $x = 5$ or $x = -2$ . Substitute both back to check: $5^2 - 3 \times 5 - 10 = 25 - 15 - 10 = 0$ , and $(-2)^2 - 3 \times (-2) - 10 = 4 + 6 - 10 = 0$ . Both hold. Why two answers? A parabola can cross the x-axis at two points, and each crossing is a solution. The craft of factoring gets its own tour in Factorising.

Two roots, lose neither

A quadratic often has two solutions, and both must be substituted back to check. Verifying only one before handing in is the classic way to lose marks.

A spare key: the quadratic formula

When factoring stalls, don't panic — there is a master key. For $ax^2 + bx + c = 0$ :

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The $\pm$ is read "plus or minus": both answers come out together. Try it on the same $x^2 - 3x - 10 = 0$ : here $a = 1$ , $b = -3$ , $c = -10$ , so

$x = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm 7}{2}$

The plus sign gives $x = 5$ , the minus sign gives $x = -2$ — same destination as factoring, by a different road.

From line to parabola

InteractiveFrom Line to Parabola

Slope m1Intercept b2

y = x + 2

y-intercept

(0, 2)

x-intercept

(-2, 0)

This is the old stage from the straight-line lesson: drag $m$ and $b$ , and for every step x grows, y grows by the same amount — which is exactly why a line never turns. Quadratics do the opposite: in the table for $y = x^2$ the steps in y are $1, 3, 5, 7$ , faster and faster, so the graph bends ever more steeply. A line moves at a constant pace, a parabola accelerates — that single bend opens a whole new world.

Check yourself

Quick quiz

1. Which one is a quadratic equation?
2. What are the solutions of (x - 5)(x + 2) = 0?
3. Using the quadratic formula on x² - 3x - 10 = 0, what is b?