How to use this applet:
This applet provides pratice for and explains the principles regarding a general form of modulus equation/inequality which you are likely to come across in your Leaving Cert. It works with problems of the same style as the example above.Select your problem using the controls provided. Clicking the 'Next' Button will bring you to the next stage. 'Random' will select a random problem for you. To get the most benefit from this applet you should try to solve the problem on paper as you go through each step. To get started why not enter the problem solved and outlined above. Enjoy!!
Notes on the maths used in the applet:
In order to solve these equations there are a few simple rules we have to know.
The modulus of x, |x|, is the positive number which has the same magnitude as x
Also
½ x ½ < k Þ -k < x < k
½ x - a ½ < k Þ -k < x - a < k
Þ a - k < x < a + k
And as we know from basic algebra squaring a negative number gives the same result as squaring the corresponding positive number.
As
|x| = x or -x
Therefore
|x|² = x²
We also must apply the rules of inequality operators which say that if you multiply or divide across by a negative number you must invert the inequality, All other operations leave it unchanged (addition, subtraction etc.) and behave as for equalities.
So if we are asked to solve a problem say:
|2x + 1| < |-x + 2|
To remove the modulus operators we square both sides. . . . . we get:
4x² + 4x + 1 < x² - 4x + 4
Bringing everything to the left hand side. . . . . we get:
3x² + 8x - 3 < 0
We factorise this. . . . .
(x + 3)(3x - 1) < 0
This tells us the roots are -3 and ¹/3
If we plot these on a number line we have three regions. To the left of -3, to the right of ¹/3 and in between them both. In general, either the region between the roots, or the regions outside the roots will satisfy the original equation. So we take three values, one from each region and test them in the original equation.
Lets choose -4, 0 and 1, choosing whole numbers makes calculations easier.
Subbing in -4 . . . . . we get
|2(-4) + 1| < |-1(-4) + 2|
|-7| < |6|
7 < 6
This is obviously false.
Simillarly subbing in 0 . . . . . we get
1 < 2
This is obviously true.
Simillarly subbing in 1 . . . . . we get
3 < 1
This is obviously false.
So in this case we see that the middle region, that is, the region between the two roots satisfies the original equation. So we write the answer as:
-3 < x < ¹/3
That is all values between - 3 and ¹/3 satisfy the equation. If the operator was <= All the values between and including would satisfy the original equation. Similarly with > and >= except in this case it would be the regions outside the roots. And finally if the operator is = it is only the two roots themselves that satisfy the original equation.
You may also encounter problems with fractions like |¹/(1 + x)| < 3 etc. These are a little trickier but are solved using a very similar technique. You square both sides. This allows you to multiply across by (1 + x)². Since it must be positive we don't have to worry about the inequality operator. Then solve exactly as outlined above.
Other useful information:
