question about inequalities and absolute values

[jc114]

I have a question about inequalities (especially in those DS questions)- since you're not allowed to cross-multiply, aren't you allowed to square each side to get rid of some x's?

For example, if (x^3) <0, then divide each side by x^2(because its always positive so no change of sign)..and then it becomes x<0?

Also, if you have absolute values in inequalities, whats the best way to get rid of them??

[Math_guru]

Lets start with the following inequality example.

Q. Consider the following inequality:
x² - 6x + 8 > 0
Solve for x.

Solution: The easiest way to solve the polynomial inequality problems without getting confused is to re-write the problem as an equality; hence we can write ---> x² - 6x + 8 = 0. Now we can factor this as (x-4)(x-2)=0

Since the problem is an inequality hence we can write (x-4)(x-2)>0 ; note that in this case since the sign is > and not >= we know that x=4 and/ or x=2 is not part of the solution. So we have to determine the boundaries/ limits of x, which can best done by using number line method. See figure 1 below.
It is clear from the number line that x is strictly > 4 and x is strictly < 2. Hence combining the two the solution is --> 2>x>4.

Ok now let us move on to Absolute value inequality:

TYPE I:

Consider, |ax+b| <= c
This can written as:

ax+b<=c AND ax+b>= - c

Now this can be solved using the above discussed methods. This also applies to polynomials.

TYPE II:

Consider, |ax+b| >= c
This can be written as

ax+b >=c OR ax+b<= - c [ Note: both are the same as the sign change causes >= to change to <= with a - sign in front of c]

Now this can be solved using the above discussed methods. This also applies to polynomials.

So, important thing to note here:

If there is <= or < in the given problem then there will 2 equations to solve for hence <= goes with AND.

If there is >= or > in the given problem then there will ONLY 1 equations to solve for hence >= goes with OR.

So lets do a problem in Absolute value inequality

Q. Solve for x in
| (x/3) - 2 | < 4/3

Solution: We see that we have a < sign so we know there will be two equations to solve for. Hence we have

(x/3) - 2 < 4/3 AND (x/3) - 2 > -4/3

(I will let you solve it)

so we get, x < 10 AND x > 2

(plot it on the number line) and the combined solution should be 2<x<10.

Hope that helps.

Good luck.

[rajesh_ctm]

That is some good stuff!!! Thanks for taking time to explain!

[diegoasaavedra]

Great explanation.
I am confuse with these, don't undestand why 4 is > and 2 is <
It is clear from the number line that x is strictly > 4 and x is strictly < 2. Hence combining the two the solution is --> 2>x>4.

Thanks

[Math_guru]

Ok, no problem in that case let me get into more details.

Inequality like this can be interpreted to find the graph of the given inequality equation, whether it is above or below the x-axis. So what is easiest place to start? Well, where the curve crosses the x-axis. Since we have already factored the above equation I am not going to do that again. Since we are solving it as an equality (to make it easier) we can say x=2 or x=4. Hence the curve crosses the x-axis at 2 and 4, and we can divide the number line into the following intervals (- &#8734;, 2) , (2,4) and (4, &#8734;). Lets say y = x² - 6x + 8 ; now pick any point in each interval. Find y at this point.

So lets start with (- &#8734;, 2), say x=0 (its easy to do the math with a 0) –
Then y = 0 – 6(0) + 8 = 8.
Thus we can say that y is always positive in the interval (- &#8734;, 2). So (- &#8734;, 2) is part of the solution as we know the given inequality is y > 0 or x² - 6x + 8 > 0

Now consider the interval (2,4) – so lets choose x=3; Then
y = (3)(3) – (6)(3) + 8 = – 1
Thus we can say that y is always negative in the interval (2,4). So (2,4) is not part of the solution as we know the given inequality is y > 0 or x² - 6x + 8 > 0

Now consider the interval (4, &#8734;) , so lets choose x=5; Then
Y = (5)(5) – 6(5) + 8 = 3
Thus we can say that y is always positive in the interval (4, &#8734;). So (4, &#8734;) is part of the solution as we know the given inequality is y > 0 or x² - 6x + 8 > 0.

Thus from above we can conclude the complete solution is x > 4 and x < 2.

Hope that helps.
Good Luck!

[diegoasaavedra]

Thanks, for your explanations and time, now i recall inequalties and the concept is clear.
Again, Thank you.

[jamesk486]

Can you also square each side to get rid of variables as well?

[Math_guru]

What do you mean by getting rid of variables? We need to find the value or the boundaries of the variable so if you cannot get rid of the variable. May be if you can give an example question it would make more sense to me. You can square to get rid of the constants but usually most of the GMAT problems don't need to do that; I haven't seen myself an inequality problem requiring to do that. Do post the question problem ok.


Hope that help!
Good Luck !

[apple100]

Lets start with the following inequality example.

Q. Consider the following inequality:
x² - 6x + 8 > 0
Solve for x.

Solution: The easiest way to solve the polynomial inequality problems without getting confused is to re-write the problem as an equality; hence we can write ---> x² - 6x + 8 = 0. Now we can factor this as (x-4)(x-2)=0

Since the problem is an inequality hence we can write (x-4)(x-2)>0 ; note that in this case since the sign is > and not >= we know that x=4 and/ or x=2 is not part of the solution. So we have to determine the boundaries/ limits of x, which can best done by using number line method. See figure 1 below.
It is clear from the number line that x is strictly > 4 and x is strictly < 2. Hence combining the two the solution is --> 2>x>4.

Ok now let us move on to Absolute value inequality:

TYPE I:

Consider, |ax+b| <= c
This can written as:

ax+b<=c AND ax+b>= - c

Now this can be solved using the above discussed methods. This also applies to polynomials.

TYPE II:

Consider, |ax+b| >= c
This can be written as

ax+b >=c OR ax+b<= - c [ Note: both are the same as the sign change causes >= to change to <= with a - sign in front of c]

Now this can be solved using the above discussed methods. This also applies to polynomials.

So, important thing to note here:

If there is <= or < in the given problem then there will 2 equations to solve for hence <= goes with AND.

If there is >= or > in the given problem then there will ONLY 1 equations to solve for hence >= goes with OR.

So lets do a problem in Absolute value inequality

Q. Solve for x in
| (x/3) - 2 | < 4/3

Solution: We see that we have a < sign so we know there will be two equations to solve for. Hence we have

(x/3) - 2 < 4/3 AND (x/3) - 2 > -4/3

(I will let you solve it)

so we get, x < 10 AND x > 2

(plot it on the number line) and the combined solution should be 2<x<10.

Hope that helps.

Good luck.

Is there another way to solve x² - 6x + 8 > 0 besides the number line method?

[Ian Stewart]

I have a question about inequalities (especially in those DS questions)- since you're not allowed to cross-multiply, aren't you allowed to square each side to get rid of some x's?

For example, if (x^3) <0, then divide each side by x^2(because its always positive so no change of sign)..and then it becomes x<0?

To answer the original question here, yes, if you have an inequality like x^3 < 0, you can divide both sides by x^2, since you know that x^2 must be positive here. That is, x^3 < 0 means the same thing as x < 0.

If you have an inequality like x^3 < x^2, you can again divide both sides by x^2 (to get x < 1). You cannot, however, divide both sides by x, since x could be positive, or could be negative; you would not know whether to reverse the inequality.

We are not, however, 'squaring both sides' here; we're just dividing both sides by a square, which is a very different thing. In general, you cannot square both sides of an inequality and expect to arrive at something which is true. For example, -3 < 2 is certainly true, but (-3)^2 is not less than 2^2. If, however, you know that everything in your inequality is positive, then you can square both sides; if 0 < a < b, it will be true that 0 < a^2 < b^2.