If you have any questions, ask them on the Beat The GMAT discussion forums. The average response time is typically __less than 30 minutes__.

- Video Course
- Video Course Overview - READ FIRST
- General GMAT Strategies - 7 videos (all free)
- Data Sufficiency - 16 videos (all free)
- Arithmetic - 38 videos (some free)
- Powers and Roots - 36 videos (some free)
- Algebra and Equation Solving - 73 videos (some free)
- Word Problems - 48 videos (some free)
- Geometry - 42 videos (some free)
- Integer Properties - 38 videos (some free)
- Statistics - 20 videos (some free)
- Counting - 27 videos (some free)
- Probability - 23 videos (some free)
- Analytical Writing Assessment - 5 videos (all free)
- Reading Comprehension - 10 videos (all free)
- Critical Reasoning - 38 videos (some free)
- Sentence Correction - 70 videos (some free)
- Integrated Reasoning - 17 videos (some free)

- Study Guide
- Office Hours
- Extras
- Guarantees
- Prices

## Comment on

Quadratic Inequalities## Brent, I read the

(1) (k – 1)(k – 3) < 0

(2) (k – 5)(k – 1)(k – 3)(k – 6) < 0

My question is - when the solution states the following:

(1) (k – 1)(k – 3) < 0 --> 1<k<3, as given that k is an integer then k=2. Sufficient.

(2) (k – 5)(k – 1)(k – 3)(k – 6) < 0 --> 5<k<6 or 1<k<3, again as given that k is an integer then k=2 (only integer IN this ranges). Sufficient.

Answer: D.

was there a way to avoid the full approach of changing to equation, finding the tow roots that =0, then using the number line, breaking it into regions and then determining the potentials solutions and whether each statement is sufficient or not? I am asking because the solution is very clear, but time wise - it is very time consuming...

## Agreed, it's very time

Agreed, it's very time-consuming. There's really no way around those calculations but, if you start with that approach immediately, you should be able to finish in under 2 minutes.

## Hi Brent,

thank you for the very useful videos.

Could you please help with the following problem from GMAT official guide:

How many of the integers that satisfy the inequality (x+1)(x+3)/(x-2)≥ 0 are less than 5?

What should I do with (x-2) in the denominator?

## First, we can see that (x+1)

First, we can see that (x+1)(x+3)/(x-2) EQUALS zero, when x = -1 and x = -3. So, we already have two integer solutions.

ASIDE: When x = 2, the expression is undefined. So, x = 2 is NOT a solution. However, x values less than 2 will make (x-2) negative, and x values greater than 2 will make (x-2) positive.

Now test the values on either side of x = -3, x = -1 and x = 2

Case A: x is less than -3

When x is less than -3, (x+3) is negative, (x+1) is negative, and (x-2) is negative. So, we get: (negative)(negative)/(negative) which evaluates to be NEGATIVE.

So, x-values that are less than -3 do NOT satisfy the inequality (x+1)(x+3)/(x-2) ≥ 0

Case B: x is greater than -3 and less than -1

There's only one integer x-value in this range (x = -2), so let's just see what we get when we plug x = -2 into the given expression.

We get: (-2 + 1)(-2 + 3)/(-2 -2) ≥ 0

Evaluate: (-1)(1)/(-4) ≥ 0

Evaluate: 1/4 ≥ 0

It WORKS, so x = -2 is another possible solution.

Case C: x is greater than -1 and less than 2

In this case, (x+3) is positive, (x+1) is positive, and (x-2) is negative. So, we get: (positive)(positive)/(negative) which evaluates to be NEGATIVE.

So, x-values BETWEEN -1 and 2 do NOT satisfy the inequality (x+1)(x+3)/(x-2) ≥ 0

Case D: x is greater than 2

In this case, (x+3) is positive, (x+1) is positive, and (x-2) is positive. So, we get: (positive)(positive)/(positive) which evaluates to be POSITIVE (PERFECT!).

So, all x-values GREATER than 2 SATISFY the inequality (x+1)(x+3)/(x-2) ≥ 0

The question asks for x-values less than 5

So, the possible integer x-values are 3 and 4.

-----------------------------------------------------------

Let's list ALL integer solutions: x = -3, x = -1, x = -2, x = 3 and x = 4

So, there are FIVE possible solutions.

## Hey Brent, I tried solving

Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

A. 0 < |x| < ½

B. |x| > ½

C. –½ < x < 0 or ½ < x

D. x < –½ or 0 < x < ½

E. x < –½ or x > 0

From the equation, we get x=0,+1/2,-1/2 when the equation is tested for =0.Now,when I plot x=+1/2 in the number line , I get all values where x>+1/2 and when I plot -1/2 in the number line, I get all values where x<-1/2.

I am highly confused on which region to consider for x=0 in the number line. Kindly help on how to progress further.

## This looks good so far.

When we plot -1/2, 0 and 1/2 on the number line, we have 4 ranges to consider:

1) x < -1/2

2) -1/2 < x < 0

3) 0 < x < 1/2

4) 1/2 < x

For range #1, plug in a value less than -1/2. How about x = -1.

When x = -1, the expression x^3 – 4x^5 = 3.

In other words, x-values in this DO NOT satisfy the given inequality.

For range #2, we can plug in x = -0.25

When x = -0.25, the expression x^3 – 4x^5 = some negative value.

In other words, x-values in this DO satisfy the given inequality.

For range #3, we can plug in x = 0.25

When x = 0.25, the expression x^3 – 4x^5 = some positive value.

In other words, x-values in this DO NOT satisfy the given inequality.

For range #4, we can plug in x = 1

When x = -0.25, the expression x^3 – 4x^5 = -3.

In other words, x-values in this DO satisfy the given inequality.

So, ranges #2 and #4 satisfy the given inequality.

Answer: C

## Thank You! Appreciate your

The technique mentioned in your tutorial does help for inequality equations eg (k – 1)(k – 3) < 0.

What would be the approach to follow when there is an expression in the denominator also. eg (k – 1)(k – 3)/(k-9) <0. In that case, can we take the approach of equating also the denominator part =0 and progress further?

## That's correct.

That's correct.

In your example, (k–1)(k–3)/(k-9) < 0, the critical points are k = 1, k = 3 and k = 9.

Once you plot those values on the number line, you will have 4 ranges to consider (just like in the previous example you asked about)

## Can you pls help explain the

Which of the following equations has 1 + √2 as one of its roots? - Meaning of the question and the concept used to solve it?

A) x^2 + 2x – 1 = 0

B) x^2 – 2x + 1 = 0

C) x^2 + 2x + 1 = 0

D) x^2 – 2x – 1 = 0

E) x^2 – x – 1= 0

## Great question.

Great question.

First of all, a "root" is the same a "solution".

So, for example, since x = 3 is the SOLUTION to the equation 2x+1=7. We can also say 3 is the ROOT of the equation 2x+1=7

We can confirm that 3 is the root by plugging 3 into the given equation.

We get: 2(3)+1 = 7

Evaluate the left side to get: 7 = 7

PERFECT!

So, one possible approach with the given question is to plug x = 1 + √2 into all 5 answer choices to see which one works.

Another approach is to recognize that the answer choices have a lot of things in common. For example, they all have an x² term.

So. let's let x = 1 + √2, and then see what x² equals

We get: x² = (1 + √2)²

= (1 + √2)(1 + √2)

= 1 + √2 + √2 + 2

= 3 + 2√2

Next, notice that 4 of the 5 answer choices have a 2x term.

So, let's let x = 1 + √2, and then see what 2x equals

We get: 2x = 2(1 + √2)

= 2 + 2√2

So, x² = 3 + 2√2, and 2x = 2 + 2√2

Since all 5 answer choices are equations that equal 0, we need to try to eliminate the terms with √2

So, if we SUBTRACT 2 + 2√2 (aka 2x) FROM 3 + 2√2 (aka x²), then we'll eliminate the terms with √2

So far, we have: x² - 2x = (3 + 2√2) - (2 + 2√2)

= 1

Since we want our equation to equal 0, we need to SUBTRACT 1 at this point.

In other words: x² - 2x - 1 = (3 + 2√2) - (2 + 2√2) - 1

= 0

Perfect! When x = 1 + √2, the equation x² - 2x - 1 equals zero.

So, the correct answer is D

Does that help?

Cheers,

Brent

## What is the logic behind this

## All 5 answer choices have an

All 5 answer choices have an x² term.

When x = 1 + √2, x² = 3 + 2√2

So, when we plug x = 1 + √2 into all 5 equations, we get:

A) (3 + 2√2) + 2x – 1 = 0

B) (3 + 2√2) – 2x + 1 = 0

C) (3 + 2√2) + 2x + 1 = 0

D) (3 + 2√2) – 2x – 1 = 0

E) (3 + 2√2) – x – 1= 0

Notice that all 5 equations (when we replace x² with 3 + 2√2) have the term 2√2

Since all 5 equations are set equal to 0, it must be the case that, when we replace the remaining x's with 1 + √2, the 2√2 must disappear.

In other words, our 2x term must be such that, when we replace x with 1 + √2, the 2√2 terms disappears.

Does that help?

Cheers,

Brent

## Hi Brent, once you have

In your explanation you need to perform the calculation 3 times with different values of x, potentially time consuming during the GMAT.

For example in the second practice question, summarised below, is there anyway to jump from stage 3 to stage 4 without going through the 3 versions of the equation?

1) x² – 8x + 9 < -6

2) x² – 8x + 15 < 0

3) (x-5)(x-3)<0

4) 3<x<5

## Question link: https:/

Question link: https://gmatclub.com/forum/if-x-is-an-integer-what-is-the-value-of-x-233...

While it's possible to create a series of rules that allow us to quickly go from Step 3 to Step 4, the rules would be somewhat convoluted (and hard to memorize!). So, I'd stick with the approach demonstrated in the video.

Please keep in mind that this question type is VERY rare. In fact, most test-takers will not encounter such a question on test day.

## Hey Brent, given that

## It would be great if we could

It would be great if we could make that generalization, but that only works for one kind of parabola (one that opens UP and has its vertex BELOW the x-axis.

Some parabolas open up and have their vertex ON the x-axis, In which case it has the form +++ 0 +++

Also some parabolas open up and have their vertex ABOVE the x-axis, In which case it has the form +++++++

Then there are times when the parabola opens DOWN and has its vertex A ABOVE the x-axis, In which case it has the form --- +++ ---

And so on.

Great idea though!

Cheers,

Brent

## Add a comment