James Gray

University of Edinburgh

Applicable Mathematics (Foundation) af0

4: Quadratics

A quadratic (polynomial) is an expression of the form

where `a`, `b` and `c` are real numbers (and `a ne 0`).

So for example `x^2 + x+2` is a quadratic as is `x^2 - 3`.

4.1 Completing The Square (H p325)

We will see later that the forms `(x+p)^2 + q` or `q-(x+p)^2` are very useful for graphing quadratics and solving quadratic equations. In this section we will consider how to rearrange a quadratic into these forms.

Example

Q: Write `x^2 + 6x` in the form `(x+p)^2 + q`.

A: First we take the coefficient of `x` and half it.  That is `6/2 = 3` in this case.

Now work out `(x+3)^2 = x^2 + 6x + 9`.  Rearranging this we get the required form

Example

Q: Write `x^2 -8x +3` in the form `(x+p)^2 + q`.

A: Again we take the coefficient of `x` and half it to get `-8/2 = -4`.

Now work out `(x-4)^2 = x^2 - 8x +16`.  So we can replace `x^2 -8x` in the quadratic above by `(x+4)^2 - 16`.  That is

  • `x^2 - 8x + 3 = (x-4)^2 - 16 +3 `
  • `\ \ \ \ \ \ \ = (x-4)^2 -13`.

Example

Q: Write `-6x^2 +30x -1/4` in the form `q - a(x+p)^2`.

A: This time we need to rearrange the quadratic in the equation before we can complete the square.  We will take out a factor of `-6` to get

  • `-6x^2 + 30x -1/4 = -6(x^2 - 5x +1/24)`

Now we complete the square for `x^2 - 5x`. 

That is we replace `x^2 - 5x` by `(x-5/2)^2 - 25/4`.  So this gives

  • `-6(x^2 - 5x +1/24) = -6((x-5/2)^2 - 25/4 +1/24)`
    • `= -6((x-5/2)^2 - 149/24)`
    • `=-6(x-5/2)^2 + 149/4`

Exercises

4.2 Solving Quadratic Equations

The solutions to `ax^2 + bx + c =0` are called the roots.

These are the places where the graph cuts the `x`-axis.  A quadratic equation may have two, one or no (real) roots. Examples of each of these possibilities are shown on the graphs below.

Completing the Square

Completing the square can always be used to solve quadratic equations.

Example

Q: Solve `x^2 - 2x -2 = 0`.

A: By completing the square `x^2 - 2x -2 = (x-1)^2 -1-2 = (x-1)^2 -3`.

To solve `x^2 - 2x -2 = 0` we use the completed square form for the left hand side to get

  • `(x-1)^2 - 3 = 0`
  • `(x-1)^2 = 3`
  • So `x -1 = +- sqrt(3)`. Therefore `x = 1 +- sqrt(3)`.

Factorising

Factorising can sometimes be used to solve quadratic equations.

Example

Q: Solve `t^2 + 4t - 12 = 0`

A: Factorise to get `(t-2)(t+6) = 0`.

So `t-2 = 0` or `t+6 = 0`.  That is `t=2` or `t=-6`.

Quadratic Formula

To solve `ax^2 + bx + c = 0`,

  • `x = (-b +- sqrt(b^2 -4ac))/(2a)`

The formula was found by using completing the square: see p149 of Heinemann.

Exercises

Solve `x^2 + 14x + 47 = 0` using both the quadratic formula and completing the square.  Which one is easier?

Solving Quadratics by:

  • completing the square: H p147 Ex 8E
    • Q2 (a), (d), (e), (h)
  • factorising: H p147 Ex 8E
    • Q1 (a), (c), (g), (i)
  • using the quadratic formula: H p150 Ex 8G
    • Q1 (a), (c), (e)

4.3 The Discriminant

For `ax^2 + bx + c =0`, we call `b^2 -4ac` the discriminant.

Notice that the discriminant is the expression that appears in the square root in the quadratic formula.

The discriminant can tell us how many roots a quadratic equation has.  There are three cases:

  • `b^2 - 4ac > 0 \ \ \ \ \ \ \ ` two real roots
  • `b^2 -4ac = 0 \ \ \ \ \ \ \ \ ` one real root
  • `b^2 -4ac < 0 \ \ \ \ \ \ \ \ ` no real roots

4.4 Graphing Quadratics

On a sketch you must show the coordinates of

  • `y`-intercept
  • `x`-intercepts (i.e. the roots)
  • turning point

The completed square forms `a(x+p)^2 +q` or `q - a(x+p)^2` tell us the following information about the graph.

Example

Q: Sketch the graph of `y=x^2 -8x +3`.

A: `y`-intercept: set `x=0`. So `y = 0^2-8 times 0 +3 =3`.

`x`-intercept: set `y=0`. So `0=x^2 - 8x +3`.  Now complete the square to solve the quadratic equation

  • `(x-4)^2 -16 +3 =0`
  • `(x-4)^2 =13`
  • `x-4 = +- sqrt(13)`
  • `x = 4 +- sqrt(13) = 0.39` or `7.61` to 2 d.p.`

We can also use the completed square form to rewrite `y=x^2 -8x +3` as `y = (x-4)^2 -13`.  Hence we know that the minimum value occurs at `(4,-13)`.

Therefore the completed graph is:

Exercises

  • Discriminant: H p150 Ex 8G
    • Q2 (a), (b), (c)
  • Discriminant: H p151 Ex 8H
    • Q1 (a), (c)
    • Q2 (b), (d)
  • Sketching the graph by completing the square: H p146 Ex 8D
    • Q1 (a), (b), (c), (d)
    • Q2 (a), (c), (e), (g)
    • Q3 (1), (2), (4)
    • Q5 (a), (d)

4.5 Quadratic Inequalities

Method:

  • Solve quadratic
  • Draw a quick sketch

Example

Q: Find the values for which `2x^2 + x -3 >0`.

A: By factorising `2x^2 + x - 3 = (2x + 3)(x-1)`.  So the solutions to `2x^2 + x -3 = 0` are `x = -3/2` or `1`.

So `2x^2 + x -3 >0`

for `x>1` or `x<-3/2`

Example

Q: Find the values for which `2x^2 + x -3 <0`.

A: Using the working and the sketch from the previous example we can see that the inequality holds when `-3/2 <x<1`.

Exercises

  • Quadratic Inequalities: H p148 Ex 8F
    • Q1
    • Q2
    • Q3 (a), (c), (e)