A quadratic equation has an unknown which appears at least once in the form x². For example: 3x² - 3x = x - 1.

We bring all the terms over to the LHS of the equation so that the RHS is equal to 0. This gives: 3x² - 4x + 1 = 0, which is the form we need to express quadratic equations in to be able to solve them. 

Once the equation is in this form it can often be simplified easily. For example:

Exercise 1.- Express the following equation in its simplest form:

3x² - 3x/2 = x/2 - x + 2 + x²

First, we find a common denominator to get rid of any denominators in the equation. This gives:

6x² - 3x = x - 2x + 4 + 2x²

Then we bring all the terms over to the LHS: 4x² - 2x - 4 = 0 and simplify (dividing by 2): 2x² - x - 2 = 0.

In the window below we are going to solve the equation  3x² - 4x + 1=0 using a graph.

Once the expression on the LHS of the equation has been simplified we are left with a quadratic function, which for the first example above is: f(x) or y=  3x² - 4x + 1

1.-Note that the graph takes the form of a curve called a "parabola".

In this example the parabola cuts the X-axis at two points. The "x" values at these points give the solution of the equation as y = 0 so: 3x² - 4x + 1 = 0 which is what we are looking for.

2.-Find these values of x by moving the point over the curve or by changing the value of x at the bottom of the window (You can also change the present x value by writing a new one in directly).

3.-Check that you get the solutions: x = 1 and x = 0.33 (which is really x = 1/3).

The solutions of the equation are also called the "roots" of the equation.

As we saw in the explanation, any quadratic equation can be expressed in the form:

ax² +bx + c = 0

where a, b and c are whole numbers (positive or negative). To get this form we may need to find a common denominator (if there are denominators in the equation) to get rid of denominators and bring all the terms over to the LHS.

Once we have got the right form, the two possible solutions of the equation are:

So, for our example equation: 3x² - 4x + 1 = 0: the solutions are:

So 1 and 0.33 are the two solutions or roots of the equation.

1.- Solve the following equations

a)( x²)/2 = x/2 + 3

b) 3x² = 12

2.-Solve both equations numerically in your exercise book ("be careful with the common denominator in the first one!").

3.-Move the red point until you find the point where the parabola cuts the X-axis. You can also change the value of x in the box at the bottom. The values of x that you find should be the same as the numerical solutions you found earlier

4.-Check that for example a) you get: "a = 1" , "b = -1" and "c = -6" which give the solutions: x = -2; x = 3.

5.-Check that for example b) the solutions are x = 2 and x = -2 ("careful!" In this case b = 0). You can probably solve this equation without using the formula (we can say that it is an incomplete quadratic equation). We just need to see that 3x² = 12 is the same as x² = 4 and therefore x = the square root of 4, which is 2 or -2. Therefore:

"If the parabola cuts the X-axis at two points, the x values at these points give the two solutions or roots of the quadratic equation".

Use the following window, changing the a, b and c parameter values appropriately, to solve the following equations graphically.

1.-Solve the following equation graphically.

a) x² - 2x -1 = 0

b) x² -1/4 = 0

c) 4x² - 4x +1 = 0

("take care with examples where the equation has only one solution or no solutions").

2.-Solve the equations numerically as well and check that the results are the same.

The following problems are based on a quadratic equation, which when solved, could in some cases, give a linear equation as an answer.

Find the hypotenuse of a right-angled triangle given that the lengths of its three sides are three consecutive numbers

Solution: Use a drawing like the one below to solve the problem, bearing in mind the length of the hypotenuse and the longest other side, given that the shortest other side has a length of "x".

Apply Pythagoras' theorem: (x+2)² = (x+ 1)²+ x².

In each of the following examples form the necessary equation to solve the problems. Find both the numerical and graphical solutions, using the window to help you.

2.- The base of a rectangle is three times its height. If we reduce the length of each side by 1 cm, the original area decreases by 15 cm². Find the original dimensions and area of the rectangle.

(Suggestion: Make a sketch of the problem).

Solution: Base = 12 cm. Height = 4 cm.

3.- Find three consecutive odd numbers which give an answer of 7 when the squares of the two smallest numbers are subtracted from the square of the biggest number.

(Solution: 5, 7 and 9 )

4.- A father's age is the square of that of his son. In 24 years time the father's age will be twice that of his son. How old are they both now?

(Solution: 6 and 36)