|
Quadratic equations. |
| 4th year of secondary education. Option A. | |
| Explanation. | |
|
An equation is a quadratic equation when the unknown is squared (x2
) at least once in the equation. E.g. 3x2 -
3x = x - 1.
If we bring all the terms over to the RHS (right hand side) and the LHS (left hand side) is equal to 0 we get: 3x2 - 4x + 1 = 0 which is the form we should always use to express quadratic equations to be able to solve them. In many cases once the equation has this form it can be simplified which is very useful. For example: Exercise 1.- Express the following equation in the simplest form possible: 3x2 - 3x/2 = x/2 - x + 2 + x2 First, we have to find a common denominator to get rid of any existing denominators in the equation. We then get: 6x2 - 3x = x - 2x + 4 + 2x2 If we bring all the terms over to the RHS we get: 4x2 - 2x - 4 = 0 and if we simplify it (by dividing it by 2) we get: 2x2 - x - 2 = 0. |
| Graphical solutions. | ||
|
Later we are going to see how to solve quadratic equations numerically but first we are going to see how to do so graphically: Once the RHS of the equation has been simplified, it becomes a quadratic function which, in the first example gives us: f(x) = 3x2- 4x + 1
| ||
|
Look at the following window to see how this is represented on a graph.
As you can see, the graph takes the form of a curve which is called a "parabola". In this example the parabola crosses the x-axis at two points. These values for x are the solutions for the equation as y=0 (i.e.) 3x2 - 4x + 1 = 0 which is what we were looking for.
| ||
|
| ||
|
Therefore:The solution of a quadratic equation are the two values given for "x" where the graph of the quadratic function (parabola) cuts the x-axis. You should have found the solutions: x = 1 and x = 0.33 (which is actually x = 1/3). The solutions of an equation are also called the "roots" of the equation. | ||
| General numerical solutions for quadratic equations. | |||
|
As we saw earlier in the explanation, any quadratic equation can be expressed with the following form: ax2 +bx + c = 0 where a, b and c are any positive or negative whole numbers. We just need to find a common denominator (if there are any denominators in the original equation) to get rid of existing denominators and then bring all the terms over to the RHS. Once the equation has this form, we know that the two "possible" solutions for the equation are:
So for the first equation we used as an example: 3x2 - 4x + 1 = 0 would give the solutions:
Therefore the solutions or roots of the equation are 1 and 0.33. | |||
|
Exercise 2 Solve the following equations: a) x2/2 = x/2 + 3 b) 3x2 = 12 Solve both equations numerically in your notebook (take care with the denominator in the first one!). Then check your answers in the following window. Read the instructions given to know what to do.
| |||
|
| |||
|
The values you get for x should be the same as the numerical solutions you got earlier. | |||
|
In example a) of the exercise you should find that: "a = 1" , "b = -1" and "c = -6" which give the following solutions to the equations: x = -2 ; x = 3. In example b) the solutions should be x = 2 and x = -2 (careful as b = 0 here!) You can probably solve this equation without using the formula (we could call it an incomplete quadratic equation). If we realize that 3x2 = 12 is the same as x2 = 4 then x = the square root of 4, (i.e.) 2 or -2.
| |||
|
Therefore: "If the parabola cuts the x-axis twice, the values of x at these points are the two solutions or roots of the quadratic equation" | |||
| Exercises. | |
|
Exercise 3
Change the values of the parameters a, b and c appropriately
in the following window to solve the following equations graphically:
a) x2 - 2x -1 = 0
b) x2-1/4 = 0
c) 4x2 - 4x +1 = 0
Then solve the equations numerically as well, in your notebook,
using the formula and check that you get the same solutions.
| |
| Solving problems with these equations. | |
|
The following problems can be solved by forming a quadratic equations. However,
sometimes you may get a simple equation when you solve them.
Problem 1. Work out the hypotenuse of a right-angle triangle given the fact that the length of the sides are three consecutive numbers. Solution: We can draw the following diagram to help. The hypotenuse is the longest side and the shortest side is "x".
If we apply Pythagoras' theorem we get: (x+2)2 = (x+ 1)2 + x2. We expand the brackets to get: x2 + 4x + 4 = x 2 + 2x + 1+ x2. If we take all the terms over to the LHS and simplify the equation we get: x2 - 2x - 3 = 0. You already know how to solve this equation numerically and graphically to get the following solutions: x = 3 and x = -1. Use the following window to solve the equation. The solution x = -1 is obviously impossible because a side cannot have a negative measurement. Therefore, we get: Hypotenuse: x + 2 = 5 ; longest other side: x + 1 = 4 ; shortest other side: x = 3.
| |
|
Work out the following problems by forming the necessary equation in each case. Solve the equations both numerically and graphically using the window above | |
|
Problem 2.
In a rectangle the base measures three times the height. If we reduce each
side by 1 cm the original area is reduced by 15 cm2. Work out the
dimensions of the original rectangle. (Suggestion: make a sketch of the problem). Solution: Base = 12 cm. height = 4 cm.
| |
|
Problem 3.
Find out which three consecutive
(Solution: 5 , 7, and 9 ). | |
|
Problem 4.
A father's age is the square of his son's age. In 24 years time the father
will be exactly twice the age of his son. How old are they now? (Solution: 6 and 36).
| |
 |
 |
|||||
 |
| Leoncio Santos Cuervo | ||
 |
||
| Spanish Ministry of Education. Year 2001 | ||
Except where otherwise noted, this work is licensed under a Creative Common License
