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## Comment on

Sum of Solutions## Alternatively, we could have

## Great approach!

Great approach!

## hey there,

are you sure we have to check for extraneous roots ??

I've never seen it and I went over gmat official guide 2017, GMAT Quantitative Review 2017, and gmat official guide 2016

They never mentioned it in their explanations

## The test-makers might not

The test-makers might not mention the term "extraneous roots," but it's indirectly implied on page 122 of the 2017 Official Guide under "5. Solving Equations by Factoring"

Here, they say that a fraction equals zero if and only if its numerator equals 0. That is, if the denominator equals zero, the fraction does not equal 0.

Another way to put it is that 2/0 ≠ 0, and 0/0 ≠ 0

## What do you think of this

Factoring the nominator we get

(x+6)(x-1)

Factoring the denominator we get:

(x-1)(x-3)

So cancel out (x-1) in the equation from nominator and denominator and we are left with:

(x+6)/(x-3)=0

this implies that the nominator should equal zero. So x+6=0 then x=-6

Is this correct?

## Perfect approach!

Perfect approach!

## Hi Brent,

Thank you for the video:

Going back to this comment

Factoring the nominator we get

(x+6)(x-1)

Factoring the denominator we get:

(x-1)(x-3)

So cancel out (x-1) in the equation from nominator and denominator and we are left with:

(x+6)/(x-3)=0

From here, I multiply (x-3) to 0, which makes (x+6)=0 -> x=-6

Is this approach correct as well?

## Yes, that approach yields the

Yes, that approach yields the correct answer. However, when you took (x+6)(x-1)/(x-1)(x-3), and "canceled" the (x-1), it's important to recognize that you were actually dividing numerator and denominator by (x-1), and when you do this, you are in danger of dividing by zero. This wasn't the case, but it's useful to be vigilant when dividing by variables.

## how about x = -3 from the

## Good question!

Good question!

Let's start here: (x + 6)(x - 1)/(x - 3)(x - 1) = 0

x = 3 is not a solution for the same reason x = 1 is not a solution.

The key reason lies in the fact that dividing by zero results in a non-number (we say k/0 is undefined)

In other words, it is NOT the case that 2/0 = 0.

Instead, we say that 2/0 is undefined.

So, let's TEST whether x = 3 works by plugging it into our equation.

We get: (3 + 6)(3 - 1)/(3 - 3)(3 - 1) = 0

Simplify to get: (9)(2)/(0)(2) = 0

Simplify more: 18/0 = 0. This is not true.

So, x = 3 is not a solution.

Cheers,

Brent

## Hi Brent,

I have two questions.

My first question is "what are the cases in which we must check for extraneous roots?"

Is the answer to this question => for "ALL quadratic equations where the solutions are requested as the answer options, ALL polynomial equations taken to the power of n where n=even number and EVERY absolute value question where the value of x is requested (when the absolute value of the term is equal to 0)"?

My second question on this matter is "Is the concept of extraneous roots tested in an INEQUALITY question type as well?" Could you please provide examples to the same if they are?

Thank you in advance!

## Check for extraneous roots

Check for extraneous roots when:

- There's an EQUATION with a variable inside a square root

- There's an EQUATION with a variable inside an absolute value

At the same time, we should always be on the lookout for inadvertently dividing by 0.

That is, we must recognize that k/0 is undefined (that is, it is not an actual number).

So, in the question above, we must recognize that x = 1 is not a solution to the following equation:

(x + 6)(x - 1)/(x - 3)(x - 1) = 0

On the GMAT, you won't be required to expand this principle to inequalities.

Cheers,

Brent

## Thank you so much for

Absolutely love your videos! Thank you for sharing them

## Thanks for the kind words!

Thanks for the kind words!

## Hi Brent,

When I solved this question, I noticed that -6 was in the solution but 1 wasn't. Could we directly pick answer choice A (-6) since 1 is not in the answer choices?

## We can plug in values to

We can plug in values to confirm that x = -6 is a solution, and that x = 1 is NOT a solution.

However, in order to conclude that the SUM of all possible solutions is -6, we must be certain that there are no other possible values of x that satisfy the equation.

For example, if x = 2 were a solution, then the sun would no longer be -6.