rrobiinn wrote:1. if (x-1)(x-2)(x^2-4)=0, what are the possible values of x?
a. -2 only b. 2 only c. 1,-2, or -4 only

2. t^2-1/t-1=2 what values may t have?
a. 1 only b. 2 only c. no value d. an infinite number of values

3. if t^2+2t/2t+4=t/2 what values may t have?
a. 2 only b. -2 only c. any value d. any value except 2 e. any value except -2

Please explain them clearly.
how can I understand that the value can be any value, no value, 'x' only, an infinite number of values or any value except 'x'?

I'm happy to help with this.

1) For the first one, we will use an idea called the Zero Product Property, which says

If A*B = 0, then either A = 0 or B = 0. By extension, if A*B*C = 0, then A = 0 or B = 0 or C = 0

The whole idea of factoring in algebra is to get a product that equals zero, so you can create separate equations, each of which equals zero. BTW, in that statement, notice that the word "or" is not a piece of garnish, but rather an essential piece of mathematical equipment.

Here, we have:
(x-1)(x-2)(x^2-4)=0
Which yields
x - 1 = 0 OR x-2 = 0 OR x^2-4 = 0

x = 1 OR x = 2 OR x^2 = 4, which means x = +/-2

Thus, x = 1 OR +2 OR -2.
Incidentally, none of the choices listed with those problem is correct.

2) PARENTHESES! PARENTHESES! PARENTHESES!
What you literally have typed is:
t^2-1/t-1=2 -----> t^2 - 1/t = 3 ----> t^3 - 1 = 3t ------> t^3 + 3t - 1 = 0
which turns out to be an unsolvable cubic equation, but what I suspect you really mean is
(t^2-1)/(t-1)=2
In this case, you can factor the numerator by the difference of square formula
https://magoosh.com/gmat/2012/gmat-quant ... o-squares/
BTW, you may find this general post on factoring helpful:
https://magoosh.com/gmat/2012/algebra-on ... to-factor/
Using the difference of two square formula, we find
[(t+1)*(t-1)]/(t-1) = 2

t+1 = 2
t = 1

Now, we have to go back and check that value in the original equation. When we do that, lo and behold, the original equation takes on an indeterminate 0/0 form, so it doesn't equal anything. Therefore, there is no solution to this question.

3) PARENTHESES! PARENTHESES! PARENTHESES!
What you have typed is:
t^2+2t/2t+4=t/2 --------> t^2 -t/2 + 5 = 0, a simple quadratic
I suspect this is not literally what you mean. I suspect what you really mean is
(t^2+2t)/(2t+4)=t/2
Factor the numerator and denominator
[t*(t+2)]/[2(t+2)] = t/2
(t/2)*[(t+2)/(t+2)] = t/2
So, the left side is t/2 times a fraction of something over itself. That second fraction, (t+2)/(t+2) will always equal 1 as long as the numerator and denominator are not zero; when this fraction equals 1, the full equation is satisfied.
When t = -2, that is, when the (t+2)/(t+2) fraction becomes the indeterminate 0/0, then we no longer can do ordinary math, and the equation no longer functions as a equation.
Thus, all real numbers except x = -2 satisfy this equation.

Any value that makes a denominator zero cannot be a solution. Any time you cancel common algebraic factors in the numerator & denominator, as part of solving, you always have to take your tentative answers back to the original equation and verify that plugging in doesn't yield something like 0/0. In fact, it's just an excellent habit in general always to take what you think is the answer to the algebra and plug it back into the original equation.

Does all this make sense?

Mike

Hi Md Raihan Uddin,

The post that you're asking about is almost 2 years old, so I'm not sure if you'll get much of a response from the original contributors. Here's the answer your question though:

Your approach: cross-multiplying, forming a quadratic = to 0 and then factoring down....DOES work. This goes to show that there are sometimes multiple ways to "do the math" and get to the correct answer. GMAT Quant questions are usually designed with multiple approaches in mind, so that the Test doesn't become a strict "only-one-appraoch-works math test."

When comparing your approach to the other approach listed, a good question to ask would be "which one is easier/faster?" In the end, that's the approach that I would want to use.

GMAT assassins aren't born, they're made,
Rich

Md Raihan Uddin wrote:Hi, Mike

Why aren't we doing no 2 in the following way?

t^2-1 = 2t-2 or t^2 -1 -2t +2 = 0
So, (t-1)^2=0 Now in this case t=1 is the answer.

Why are we putting back the value of t in original question?
So far I know, we have to check back equation only in case of Mod.
So, why are we doing the same here?

Dear Md Raihan Uddin,
I'm happy to respond.

First of all, with all due respect, you made precisely the same parenthesis mistakes as the original poster on this thread. Please read:
https://magoosh.com/gmat/2013/gmat-quant ... g-symbols/

Checking your answer is a good practice in general. Don't think of this as an procedure of limited utility: it's always a good idea if you have time. There are SEVERAL problem types in which you have to check your answer. These include
1) absolute value equations, which some folks call "mods"
2) equations with square roots
3) fraction with algebraic expressions in the denominator.
Basically, in any situation in which something mathematically illegal could happen (square root of a negative, divide by zero, etc.) you have to check to make sure that the answer you found, the answer given by legitimate algebraic procedures, doesn't cause something illegal to happen.

Here, we do perfectly correct algebra, as you did, to find x = 1, but then notice that this ostensibly perfectly acceptable solution produces a "divide by zero" situation in the original equation, so this it doesn't work.

Does all this make sense?
Mike

Hi Md Raihan Uddin,

I was so keen on double-checking your Quadratic approach that I didn't refer back to the ORIGINAL question.

Since the denominator of the original fraction is (T-1), there is a value that does NOT provide a real-number solution. If T = 1, then the denominator = 0, which leads to a result that is NOT a real number. In the end, the one solution to the Quadratic (T=1) is actually does NOT provide a solution to the overall equation.

GMAT assassins aren't born, they're made,
Rich