Question: Sum of Solutions

1 is not a possible solution, since we get 0/0 = 0, if we plug x=1 into the equation.

We say "possible" because many quadratic equations have two solutions. For example, the equation x² - 5x + 6 = 0, has two possible solutions, That is, we can conclude that either x = 2 or x = 3. In this case, the sum of the possible solutions is 5 (2+3)

The video question is a bit of a trick question, because there's only 1 solution. So, it's a but misleading to say "sum of ALL possible solutionS."

That said, the word "possible" doesn't change the meaning of the question.

Aside: plugging in the answer choices only works if we are certain that there's only ONE correct solution. If there were TWO correct solutions, then plugging in the answer choices would not help.

When there are variables in the denominator (as well as in the numerator), we must be careful to confirm that the denominator never equals zero.

That technique works for this question, but could get you into trouble with other questions.

The truth of the matter is that (x² + 5x - 6)/(x² - 4x + 3) does not equal (x + 6)/(x - 3). We know this, because if we let x = 1, then the two quantities are not equal.

Sorry, but I don't think there's a faster approach.

That said, checking the two solutions for extraneous roots might take a while, but this question can still be solved in the required time.

You're right in that we don't know whether x^2 - 4x + 3 is positive. However, that doesn't matter when we're dealing with equations (EDIT: ...and quantitative comparison questions).

That only matters when we're dealing with inequalities.

More here: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

And here: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

Whenever we multiply by a variable (or variable expression), we must keep in mind that we MAY BE inadvertently multiplying by zero.

Take, for example, this easier equation: (x)(x-1)/x = 0
If we multiply both sides by x we get: (x)(x-1) = 0, which means either x = 0 or x = 1

However, we cannot say that x = 0 and x = 1 are BOTH solutions to the original equation. More specifically, we can't say that x = 0 is a solution to the original equation (x)(x-1)/x = 0, because when we replace x with 0, we end up getting 0/0 = 0, but 0/0 does not equal 0. 0/0 is undefined.

For the question in the video, we are asked to find solutions to the equation (x²+5x-6)/(x²-4x+3) = 0.

So, even though we were able to create a new (and possibly faulty) equation by multiplying both sides by (x²-4x+3), we cannot ignore the fact that were are asked to find solutions to the original equation.

Great questions, Deepak.

Yes, if you simplify an equation by multiplying or dividing each side by a variable expression, you must be sure to check the original equation to ensure you don't any extraneous roots.

NOTE: This strategy also applies to:
- equations involving square roots. More here: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

- equations involving absolute value. More here: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

As for your other questions...

1. If you're going to multiply or divide both sides of an inequality by a variable expression, then you must be sure to determine whether the expression is positive or negative. If the expression can be either positive or negative, then you must consider both cases (yes, it's the same for Quantitative Comparison (QC) questions on the GRE)

2. Yes, you must not forget the original equation.
There's a small error in your solution. Before I get to it, let's cover an important concept:

CONCEPT: If x/y = 0, then it must be the case that x = 0
In other words, we need only consider the numerator.
We cannot conclude that y = 0, because x/0 is undefined.
In fact, if x/y = 0, we can be certain that y ≠ 0.

So, once you get to (x+6)(x-1)/(x-3)(x-1) = 0, we need only focus on the numerator to conclude that (x+6)(x-1) = 0, which means the only POSSIBLE solutions are x = -6 and x = 1

Also, notice that if (x+6)(x-1)/(x-3)(x-1) = 0, we can also conclude that (x-3)(x-1) ≠ 0. In other words, we know that x CANNOT equal 3, and x CANNOT equal 1.

I mention this because you suggested that x = 3 might be a possible solution, when it is not a possible solution, since something/0 ≠ 0

Cheers,
Brent

Thanks for the kind words!
I have added extra time to your account.

Cheers,
Brent

Glad to hear!

Hi Ravi,

The fundamental issue with that solution is that fractions with zero in the denominator, are not actual numbers.
For example, we cannot say that 6/0 = 0, and we can't say that 0/0 = 0

We say that 6/0 is UNDEFINED, means it's not an actual number.
Here's why.
First notice that, since 6/2 = 3, we can also conclude that (2)(3) = 6
Likewise, since 20/4 = 5, we can also conclude that (4)(5) = 20

For a moment, let's say k is a number such that 6/0 = k
This would also imply that (0)(k) = 6
Since no value of k can satisfy that equation, we can conclude that 6/0 is undefined.
Likewise, 0/0 is undefined.

If we take the given equation: (x² + 5x - 6)/(x² - 4x + 3) = 0

...and factor numerator and denominator we get: (x + 6)(x - 1)/(x - 3)(x - 1) = 0

The solutions APPEAR to be: x = -6, x = 1 and x = 3

However, when we plug these values back into the original equation, we find that two of them DO NOT satisfy the equation.

When x = -6, we get: ((-6)² + 5(-6) - 6)/((-6)² - 4(-6) + 3) = 0
This evaluates to get: 0/63 = 0
So, x = -6 IS a solution.

When x = 1, we get: (1² + 5(1) - 6)/(1² - 4(1) + 3) = 0
This evaluates to get: 0/0 = 0
Since 0/0 is NOT an actual number, we can't say that 0/0 = 0
So, x = 1 is NOT a solution.

When x = 3, we get: (3² + 5(3) - 6)/(3² - 4(3) + 3) = 0
This evaluates to get: 18/0 = 0
Since 18/0 is NOT an actual number, we can't say that 18/0 = 0
So, x = 3 is NOT a solution.

Does that help?

Your approach is perfect.
The primary mistake that many students make is incorrectly concluding that x = 1 and x = 3 are also solutions to the equation.
You didn't fall into that trap :-)