If x = a + 1 and b = x + 2 then which of the following must be false?

why (I) is wrong !?
a=x-1,b=x-2
clearly, a>b

I think D would be the answer; a = x-1 and b = x-2 are provided... now one method to find the answer is to plug values; the other method is method of elimination so we will get a-b = 1...

Sent from my GT-I9060I using GMAT Club Forum mobile app

My approach was a hybrid:

I) & II) were straightforward FALSE/TRUE. -> Option D was ruled out
III) Didn't even made any calculation since was present in all the remaining answers
IV) Did the math
a=b^2 ; x-1 = (x+2)^2 ; x^2+3x+5=0 ; since I couldn't find any "integer" solution I flagged it FALSE
V) Did the math
b=a^2 ; x+2 = (x-1)^2 ; x^2-3x+1=0; since I couldn't find any "integer" solution I flagged it FALSE

Wrongly, I selected choice E

Making the assumption of the integer solution I didn't think of any other real number as roots of the equations on IV and V.

Thanks for this good question stonecold, even knowing that real numbers could be a possible solution, I wouldn't figure out to use the Discriminant of a Quadratic to help me out.

Ans C.
Given, x= a+1 and b = x + 2, or b = a +1 +2 = a+3
So, we have b = a + 3
Now check all the cases
1. a> b -- False
2. b >a -- True
3. a = b --False
4. a = b^2
substitute in the final equation and you will get
b^2 -b +3 = 0, imaginary roots ( remember b^2-4ac) --- False
5. b = a^2
Same as above substitute and you will get a^2 -a -3 = 0, it has real roots --- True
Hence 1, 3 and 4 are false.

[quote="Sumo23"]I think D would be the answer; a = x-1 and b = x-2 are provided... now one method to find the answer is to plug values; the other method is method of elimination so we will get a-b = 1...

What if a = -1/2 then option 1 is correct

I think a very tough ques

There are many issues with this question - GMAT Roman Numeral questions always have exactly three items (not the five you see here), there is no reason to even look at item III because it's in every answer choice, and it's bizarrely convoluted to ask which items must be false instead of which could be true.

We know a = x-1 and b = x + 2. Since x+2 is exactly 3 greater than x-1, certainly b > a. So I and III must be false.

We know b is precisely 3 greater than a, so b = a + 3. If we want to see if IV can be true, we can then ask if it's possible for a = (a + 3)^2. Notice this won't be true if a+3 is a 'fraction' between 0 and 1, because then a is negative, so could never equal a square, which is positive. But if a+3 is not between 0 and 1 (inclusive), then (a+3)^2 is always greater than a+3, which is always greater than a. So (a+3)^2 is always larger than a, and b^2 can never equal a, so IV must be false.

On the other hand, it's definitely going to be possible for a^2 = a + 3 to be true, for two values of a. If you picture y = x^2 and y = x + 3 in the coordinate plane, you can see immediately that the line y = x + 3 will meet the parabola y = x^2 in two points. Or you can notice that when a = 2, a^2 < a + 3, but when a = 3, a^2 > a + 3, so for some value of a between 2 and 3 they're equal.

But this is not a realistic GMAT question at all.

I'd add, just in case any test takers are hoping to find more information about some of the more algebraic solutions in this thread, that some of the posts above mean to discuss the "discriminant" of a quadratic (not the "determinant", which is a term used in matrix algebra, but not something you'd ever even have occasion to think about when doing GMAT questions).

Long Question.....maybe too long for the GMAT? But great question for practice

Which of the following MUST be False?

That means if we can prove that a Statement could POSSIBLE be True, then we have to Eliminate it.

I, II, and III deal with the relationship of A vs. B

X = A + 1
A = X - 1

and

B = X + 2


I. A > B
Is: X -1 > X + 2 ?

Is: X > X + 3?

can NEVER be True ---- KEEP I


II. B > A

Is: X + 3 > X?

Is: 3 > 0?

NOT Always False ---- Eliminate II


III. B = A

Is: 3 = 0?

can NEVER be True ---- KEEP III



IV. A = (B)^2

X - 1 = (X + 2)^2
X - 1 = (X)^2 + 4X + 4

(X)^2 + 3X + 5 = 0


Rule: using the Discriminant, we can figure out if there are any Root Solutions that Solve the Quadratic

(b)^2 - 4ac > 0 --------> 2 Root Solutions

(b)^2 - 4ac = 0 -------> 1 Root Solution

(b)^2 - 4ac < 0 ------> there are NO ROOT Solutions to the Quadratic

b = 3
a = 1
c = 5

(3)^2 - (4 * 1 * 5) =
9 - 20 = -11

since the Discriminant is < 0 ------> there are NO Root Solutions and IV can NEVER BE TRUE

KEEP IV


IV. B = (A)^2

X + 2 = (X - 1)^2

using the Same Concept of the Discriminant, you will find that there is a possible Root Solution to the Ultimate Quadratic. Therefore, it is possible for B = (A)^2


I ----- III ------ IV -------> can NEVER BE True and therefore MUST BE FALSE


C

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.