Quadratic Equation

An equation in which the highest power of the variable is 2 is a quadratic equation (from the Latin quadratus, “square”: the variable is squared). The general quadratic equation in one variable is

$$ax^2 + bx + c = 0 \qquad (a \ne 0)$$

where a, b and c are real numbers. If a were 0, there would be no $x^2$ term, and the equation would be linear. So $x^2 - 9 = 0$, $2x^2 + 3x = 0$ and $5t^2 - 20t + 15 = 0$ are quadratic equations, but $x^3 - x = 0$ and 4x − 1 = 0 are not.

The same expression made into a rule, $y = ax^2 + bx + c$, is a quadratic function: each x gives one y. Its graph is a curve shaped like ∪ or ∩, called a parabola. Where it is placed, and how wide it is, depend on the numbers a, b and c. The point where a parabola stops going down and starts going up (or the other way round) is its turning point, also called its vertex.

Quadratic equations are much older than algebra’s name: clay tablets from Babylon, written around 1800 BC, already solve problems about areas by a method that is, in effect, completing the square.

The form $y = ax^2 + bx + c$ hides the vertex. The form $y = a(x - h)^2 + k$, called the vertex form, shows it at once. Take $y = (x - 2)^2 - 1$. A square is never negative, so $(x - 2)^2$ is 0 or more, and it is 0 only when x = 2. So y is never less than −1, and it is exactly −1 at x = 2: the lowest point, the vertex, is (2, −1).

To turn $ax^2 + bx + c$ into vertex form we complete the square: we make the $x^2$ and x terms into the start of a perfect square. Since

$$(x + 3)^2 = x^2 + 6x + 9$$

the expression $x^2 + 6x$ is a perfect square with 9 missing. And 9 is the square of half of 6. So

$$\begin{aligned} x^2 + 6x + 5 &= (x^2 + 6x + 9) - 9 + 5 \\ &= (x + 3)^2 - 4 \end{aligned}$$

Rule: add the square of half the coefficient of x, and take it away again. Adding and taking away the same number changes nothing, but it lets the first three terms close up into a square.

The name comes from a picture, as al-Khwarizmi drew it. Draw $x^2$ as a square of side x, and split 6x into two strips, each x long and 3 wide, along two sides of it. The L-shape is almost a square of side x + 3: only a corner of 3 by 3 is missing. Filling that corner, adding 9, completes the square.

When the coefficient of $x^2$ is not 1, take it out as a factor first:

$$2x^2 + 12x + 7 = 2(x^2 + 6x) + 7 = 2\{(x + 3)^2 - 9\} + 7 = 2(x + 3)^2 - 11$$

The general case

The same steps work for any $y = ax^2 + bx + c$. Take a out, then complete the square with half of b/a, which is b/(2a):

$$\begin{aligned} y &= ax^2 + bx + c \\ &= a\left\{x^2 + \frac{b}{a}x + \frac{c}{a}\right\} \\ &= a\left\{x^2 + 2 \cdot x \cdot \frac{b}{2a} + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a}\right\} \end{aligned}$$

The first three terms are the square of x + b/(2a):

$$\begin{aligned} y &= a\left\{\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a}\right\} \\ &= a\left\{\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right\} \\ &= a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a} \end{aligned}$$

So

$$y = a\left\{x - \left(-\frac{b}{2a}\right)\right\}^2 + \frac{4ac - b^2}{4a} \quad \ldots \text{(i)}$$

Comparing (i) with the vertex form $y = a(x - h)^2 + k$ … (ii):

$$h = -\frac{b}{2a}, \qquad k = \frac{4ac - b^2}{4a}$$

The vertex of the parabola $y = ax^2 + bx + c$ is

$$(h, k) = \left(-\frac{b}{2a},\ \frac{4ac - b^2}{4a}\right)$$

and its axis of symmetry is the line x = −b/(2a). In practice, k is usually quicker found by putting x = h into the function.

A sketch is a graph drawn from its important points, not from dozens of plotted ones. For a parabola $y = a(x - h)^2 + k$, four properties give those points.

1. Shape. If a > 0, the parabola opens upward (∪): the vertex is its lowest point (minimum), and it has no highest point, since it goes on rising for ever. If a < 0, it opens downward (∩): the vertex is its highest point (maximum), and it has no lowest point. The larger a is (ignoring its sign), the narrower the parabola.

2. Vertex. The vertex is (h, k).

3. Intercepts.

   • a.The y-intercept: put x = 0. For $y = ax^2 + bx + c$ it is simply c; in vertex form it is $ah^2 + k$, the same number.
   • b.The x-intercepts: put y = 0 and solve $a(x - h)^2 + k = 0$. If its roots are x₁ and x₂, the parabola cuts the X-axis at (x₁, 0) and (x₂, 0). A parabola may cut the X-axis twice, touch it once, or miss it.

4. Axis. The axis of symmetry is the upright line x = h through the vertex. The parabola is its own mirror image in this line, so every point on one side has a partner at the same height on the other. That gives one extra point for free: the y-intercept’s partner.

Steps for sketching y = ax² + bx + c

1. Write it in vertex form; read the vertex and the axis.
2. Say which way it opens, from the sign of a.
3. Find the y-intercept, and its mirror point across the axis.
4. Find the x-intercepts, if any (factorize, or take square roots of the vertex form).
5. Plot these points and draw a smooth curve through them, round at the vertex (never a point like a V).

Worked examples

Every parabola $y = (x - h)^2 + k$ is the parabola $y = x^2$ slid to a new place. Sliding every point of a shape the same distance in the same direction is a translation, as in 9.2 Translation. A parabola can be translated horizontally (left or right), vertically (up or down), or both, a combined translation. Its size and shape don’t change.

1. Horizontal translation

Why does a minus 2 move the graph to the right? Because $y = (x - 2)^2$ does everything $y = x^2$ does, 2 units later. $y = x^2$ reaches the height 4 at x = 2; $y = (x - 2)^2$ reaches it only when x − 2 = 2, at x = 4. Every height comes 2 further along: the whole curve has moved right.

When $y = x^2$ is translated h units to the right, it becomes $y = (x - h)^2$; translated h units to the left, it becomes $y = (x + h)^2$. The parabola keeps its size and shape.

2. Vertical translation

In the same way, $y = x^2$ moved 2 units down is $y = x^2 - 2$. The vertex moves from (0, 0) to (0, 2) or (0, −2).

When $y = x^2$ is translated k units up, it becomes $y = x^2 + k$; translated k units down, it becomes $y = x^2 - k$. Its size and shape stay the same. Here the sign does what you would expect: adding k to every height lifts the curve.

3. Combined translation

When $y = x^2$ is translated h units to the right and then k units up, it becomes $y = (x - h)^2 + k$, with its vertex at (h, k) and its axis x = h. This is why $y = (x - h)^2 + k$ is called the vertex form of a parabola. (A negative h means left, a negative k down.)

The number a in $y = a(x - h)^2 + k$ does one more thing before the slide: it stretches $y = x^2$ upward (a > 1: narrower), squashes it (0 < a < 1: wider), and, when negative, turns it upside down. Try every case:

Here is every case in one table:

No.	Equation of the parabola	Vertex	Opens
1	$y = x^2$	(0, 0)	upward
2	$y = -x^2$	(0, 0)	downward
3	$y = ax^2$, a > 0	(0, 0)	upward
4	$y = -ax^2$, a > 0	(0, 0)	downward
5	$y = x^2 + k$	(0, k)	upward
6	$y = -x^2 + k$	(0, k)	downward
7	$y = (x + k)^2$	(−k, 0)	upward
8	$y = (x - k)^2$	(k, 0)	upward
9	$y = a(x - h)^2 + k$	(h, k)	upward if a > 0, downward if a < 0

Question 1

Question 2

Convert each quadratic function into the vertex form $y = a(x - h)^2 + k$ and sketch its graph.

Question 3

For each quadratic equation: i. write it in vertex form; ii. find the vertex of its parabola; iii. find the equation of the axis of symmetry; iv. show its graph as a translation of $y = x^2$.

The general quadratic equation is $ax^2 + bx + c = 0$, with a, b, c real numbers and a ≠ 0. It can be solved by factorizing, by the formula, by completing the square, and also by drawing a graph. The graph shows what the solutions are.

Two ways of reading an equation as a graph:

• One curve and the X-axis. The solutions of $ax^2 + bx + c = 0$ are the values of x where $y = ax^2 + bx + c$ is 0, so they are the x-coordinates of the points where the parabola cuts the X-axis: its x-intercepts.
• A parabola and a line. Rearrange the equation as $x^2 =$ (something linear), dividing by a first if a ≠ 1. The solutions are where $y = x^2$ and that straight line meet: there, both sides of the equation are equal. Only one parabola, $y = x^2$, is ever needed, and the line is quick to draw.

Either way, two meeting points give two solutions, a line or axis that only touches the curve gives one (a repeated root), and none gives no real solution.

Worked examples

A quadratic function $y = ax^2 + bx + c$ and a linear equation such as y = mx + d together are a pair of equations. A solution of the pair is a point (x, y) that fits both: a point on the parabola that is also on the line. So, to solve them by graph, plot both on the same axes and read the points where the parabola and the line intersect. Here the answer is a point, both coordinates.

Worked examples

To find a parabola’s equation from its graph, go backwards: put the coordinates of points read off the graph into $y = ax^2 + bx + c$, and solve for a, b and c. Three points give three equations; a point on the Y-axis gives c at once.

Back to the ball

Question 1

Solve each equation by the graphical method.

Question 2

Solve each pair of equations by the graphical method.

Question 3

Find the equation of each parabola.

In the SEE’s Optional Mathematics paper (75 marks, 3 hours), Quadratic Equation is one of the units of Algebra (Units 1 to 6 of this book). CDC’s specification grid gives Algebra 17 marks, in three kinds of question:

Group	Kind of question	Algebra’s marks	Levels
A	multiple choice, 1 mark each	2	knowledge 1, application 1
B	within the content area: questions on Algebra alone, each of 2 to 4 parts	10	knowledge 2, understanding 2, application 4, higher ability 2
C	across content areas: one question joining Algebra with another area	5	understanding 2, application 2, higher ability 1

In CDC’s model paper this unit has a 1-mark multiple-choice question in group A, at the knowledge level: what the real roots of $ax^2 + bx + c = 0$ are on its graph. And a group B question starting from a composite function (Unit 1) goes on to quadratics: convert a quadratic function into the vertex form $y = (x - h)^2 + k$ (1 mark, understanding), then describe the combined translation that takes $y = x^2$ to it (2 marks, higher ability: one mark for each move with its direction and size).

What each kind asks, and where its marks are:

• Vertex form: completing the square, the vertex and the axis read off.
• Describing the translation: h units right or left, then k units up or down, each with its direction.
• Sketching: the shape, vertex, intercepts and axis, then a smooth curve.
• Solving by graph: a table of values, the curve (and line) drawn, and the solutions read off the meeting points.
• The equation from a graph: points into $y = ax^2 + bx + c$, the three equations solved.

Here is a question of the longer kind, written for this book, with how it is marked.