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## 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

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