Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 28 May 2017, 14:41

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If y= (x-1)(x+2), then what is the least possible value of y

Author Message
TAGS:

### Hide Tags

Director
Joined: 29 Nov 2012
Posts: 885
Followers: 15

Kudos [?]: 1194 [0], given: 543

If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

07 Jul 2013, 07:39
15
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

48% (02:16) correct 52% (01:17) wrong based on 233 sessions

### HideShow timer Statistics

If y= (x-1)(x+2), then what is the least possible value of y?

A. -3
B. -9/4
C. -2
D. -3/2
E. 0

Any alternative solutions?
[Reveal] Spoiler: OA

_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1122
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 191

Kudos [?]: 2100 [5] , given: 219

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

07 Jul 2013, 07:55
5
KUDOS
1
This post was
BOOKMARKED
fozzzy wrote:
If y= (x-1)(x+2), then what is the least possible value of y?

A. -3
B. -9/4
C. -2
D. -3/2
E. 0

Any alternative solutions?

The function is a parabola with a positive "a" coefficient

The roots are x=1 and x=-2, for values between 1 and -2 it will have negative values (E is out)

$$y=x^2+x-2$$, you can try to insert values (starting from the least) to see if it could be the answer. Example
$$-3=x^2+x-2$$ or $$x^2+x+1=0$$ =>Impossible
$$-\frac{9}{4}= x^2+x-2$$ or $$x^2+x+\frac{1}{4}=0$$ => $$x=-0.5$$ valid => CORRECT

Or approach #2:

Given the roots -2 and 1, the x of the vertex will be the middle point => $$x=-0.5$$

And the least value will be the y coordinate of the vertex (plug $$x=-0.5$$ into the equation).
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Math Expert
Joined: 02 Sep 2009
Posts: 39037
Followers: 7750

Kudos [?]: 106471 [6] , given: 11626

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

07 Jul 2013, 07:58
6
KUDOS
Expert's post
8
This post was
BOOKMARKED
fozzzy wrote:
If y= (x-1)(x+2), then what is the least possible value of y?

A. -3
B. -9/4
C. -2
D. -3/2
E. 0

Any alternative solutions?

$$y= (x-1)(x+2)=x^2+x-2$$.

Theory:
Quadratic expression $$ax^2+bx+c$$ reaches its extreme values when $$x=-\frac{b}{2a}$$.
When $$a>0$$ extreme value is minimum value of $$ax^2+bx+c$$ (maximum value is not limited).
When $$a<0$$ extreme value is maximum value of $$ax^2+bx+c$$ (minimum value is not limited).

You can look at this geometrically: $$y=ax^2+bx+c$$ when graphed on XY plane gives parabola. When $$a>0$$, the parabola opens upward and minimum value of $$ax^2+bx+c$$ is y-coordinate of vertex, when $$a<0$$, the parabola opens downward and maximum value of $$ax^2+bx+c$$ is y-coordinate of vertex.

Examples:
Expression $$5x^2-10x+20$$ reaches its minimum when $$x=-\frac{b}{2a}=-\frac{-10}{2*5}=1$$, so minimum value is $$5x^2-10x+20=5*1^2-10*1+20=15$$.

Expression $$-5x^2-10x+20$$ reaches its maximum when $$x=-\frac{b}{2a}=-\frac{-10}{2*(-5)}=-1$$, so maximum value is $$-5x^2-10x+20=-5*(-1)^2-10*(-1)+20=25$$.

Back to the original question:
$$y= (x-1)(x+2)=x^2+x-2$$ --> y reaches its minimum (as $$a=1>0$$) when $$x=-\frac{b}{2a}=-\frac{1}{2}$$.

Therefore $$y_{min}=(-\frac{1}{2}-1)(-\frac{1}{2}+2)=-\frac{9}{4}$$

Or:

Use derivative: $$y'=2x+1$$ --> equate to 0: $$2x+1=0$$ --> $$x=-\frac{1}{2}$$ --> $$y_{min}=(-\frac{1}{2}-1)(-\frac{1}{2}+2)=-\frac{9}{4}$$

Hope it's clear.
_________________
Director
Joined: 29 Nov 2012
Posts: 885
Followers: 15

Kudos [?]: 1194 [2] , given: 543

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

07 Jul 2013, 08:04
2
KUDOS
Here's how I did this one...

$$x^2 + x - 2$$

we can complete the square since $$(x+a)^2 = x^2 + 2ax + a^2$$

Here $$2ax = x$$ >>$$a=\frac{1}{2}$$

$$(x+1/2)^2 - 1/4 - 2 = y$$ subtracting 1/4 since 1/4 will be added when its squared.

$$(x+1/2)^2 - 9/4 = y$$

$$y = -9/4$$
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 629
Followers: 83

Kudos [?]: 1194 [0], given: 136

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

07 Jul 2013, 10:41
fozzzy wrote:
If y= (x-1)(x+2), then what is the least possible value of y?

A. -3
B. -9/4
C. -2
D. -3/2
E. 0

Any alternative solutions?

$$y = x^2+x-2 \to x^2+x-(2+y)=0 \to$$ For real values of x, the Discriminant$$(D)\geq{0} \to 1^2-4*1*[-(2+y)]\geq{0}\to$$

$$y+2\geq{-\frac{1}{4} }\to y\geq{-\frac{9}{4}}$$

B.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15501
Followers: 651

Kudos [?]: 210 [0], given: 0

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

27 Oct 2014, 22:04
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.
_________________
Manager
Joined: 05 Jun 2014
Posts: 72
GMAT 1: 630 Q42 V35
Followers: 1

Kudos [?]: 26 [0], given: 51

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

28 Oct 2014, 13:08
Can we also use b^2 - 4ac to determine the valid minimum value, as thats what I did.. I got 0 as the answer for B and I thought 0 means invalid so went for C. I want some clarification here, first if b^2-4ac=0, it means eq has no real soln?? Or my approach for this question was wrong ! Thanks in advance
Manager
Joined: 30 Mar 2013
Posts: 136
Followers: 0

Kudos [?]: 42 [0], given: 196

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

29 Oct 2014, 02:09
I did it this way
The minimum (or max, if the coef is -a) should lie half way between the two roots which are -2 and 1. halfway between that is -3/2.
Now put in -3/2 in place of x:
(-3/2 -1)(-3/2 +2) = -9/4
Current Student
Joined: 26 Aug 2014
Posts: 827
Concentration: Marketing
GPA: 3.4
Followers: 7

Kudos [?]: 160 [0], given: 98

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

29 Oct 2014, 03:40
can someone tell me why it is wrong to just test all the answer values for for x and find the lowest number? That would get you to answer E (-2).

I missed a question once because I assumed that a y= (x equation). Meant x equation was equal to zero. When do we assume it's a quadratic equation?
Manager
Joined: 30 Mar 2013
Posts: 136
Followers: 0

Kudos [?]: 42 [0], given: 196

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

29 Oct 2014, 13:57
When X has a power 2. Quad comes from the word "square"
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15501
Followers: 651

Kudos [?]: 210 [0], given: 0

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

17 Mar 2016, 21:38
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.
_________________
Manager
Joined: 24 May 2013
Posts: 86
Followers: 0

Kudos [?]: 26 [0], given: 99

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

18 Mar 2016, 06:37
f(x) = (x-1)(x+2) =x^2+x-2

Taking derivative
F'(x) = 2x+1 =0 =>x=-1/2
f(x) = 1/4 -1/2 -2 = -9/4

Hence B
Director
Joined: 24 Nov 2015
Posts: 562
Location: United States (LA)
Followers: 11

Kudos [?]: 30 [0], given: 227

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

04 Jul 2016, 03:58
Bunuel wrote:
fozzzy wrote:
If y= (x-1)(x+2), then what is the least possible value of y?

A. -3
B. -9/4
C. -2
D. -3/2
E. 0

Any alternative solutions?

$$y= (x-1)(x+2)=x^2+x-2$$.

Theory:
Quadratic expression $$ax^2+bx+c$$ reaches its extreme values when $$x=-\frac{b}{2a}$$.
When $$a>0$$ extreme value is minimum value of $$ax^2+bx+c$$ (maximum value is not limited).
When $$a<0$$ extreme value is maximum value of $$ax^2+bx+c$$ (minimum value is not limited).

You can look at this geometrically: $$y=ax^2+bx+c$$ when graphed on XY plane gives parabola. When $$a>0$$, the parabola opens upward and minimum value of $$ax^2+bx+c$$ is y-coordinate of vertex, when $$a<0$$, the parabola opens downward and maximum value of $$ax^2+bx+c$$ is y-coordinate of vertex.

Examples:
Expression $$5x^2-10x+20$$ reaches its minimum when $$x=-\frac{b}{2a}=-\frac{-10}{2*5}=1$$, so minimum value is $$5x^2-10x+20=5*1^2-10*1+20=15$$.

Expression $$-5x^2-10x+20$$ reaches its maximum when $$x=-\frac{b}{2a}=-\frac{-10}{2*(-5)}=-1$$, so maximum value is $$-5x^2-10x+20=-5*(-1)^2-10*(-1)+20=25$$.

Back to the original question:
$$y= (x-1)(x+2)=x^2+x-2$$ --> y reaches its minimum (as $$a=1>0$$) when $$x=-\frac{b}{2a}=-\frac{1}{2}$$.

Therefore $$y_{min}=(-\frac{1}{2}-1)(-\frac{1}{2}+2)=-\frac{9}{4}$$

Or:

Use derivative: $$y'=2x+1$$ --> equate to 0: $$2x+1=0$$ --> $$x=-\frac{1}{2}$$ --> $$y_{min}=(-\frac{1}{2}-1)(-\frac{1}{2}+2)=-\frac{9}{4}$$

Hope it's clear.

Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them
Math Expert
Joined: 02 Sep 2009
Posts: 39037
Followers: 7750

Kudos [?]: 106471 [1] , given: 11626

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

04 Jul 2016, 04:02
1
KUDOS
Expert's post
rhine29388 wrote:
Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them

If you know how to use any of the advanced techniques of course you can use it. Personally I've never seen a GMAT question requiring something like this. For example, I've never seen a geometry question requiring trigonometry, every GMAT geometry question can be solved without it.
_________________
Director
Joined: 24 Nov 2015
Posts: 562
Location: United States (LA)
Followers: 11

Kudos [?]: 30 [0], given: 227

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

04 Jul 2016, 07:03
Bunuel wrote:
rhine29388 wrote:
Can we use derivatives and integration to solve such type of problems? would that a bit advanced as per the Gmat exam,however if it helps i don't see the problem in using them

If you know how to use any of the advanced techniques of course you can use it. Personally I've never seen a GMAT question requiring something like this. For example, I've never seen a geometry question requiring trigonometry, every GMAT geometry question can be solved without it.

Always a pleasure learning from you kudos to your post
Manager
Joined: 26 Dec 2015
Posts: 116
Location: United States (CA)
Concentration: Finance, Strategy
WE: Investment Banking (Venture Capital)
Followers: 0

Kudos [?]: 23 [0], given: 1

Re: If y= (x-1)(x+2), then what is the least possible value of y [#permalink]

### Show Tags

05 Jul 2016, 11:52
i did NOT do this problem like the other posters did on here...way too complicated. i just plugged in #s.

start w/ x=1. if x=1, (1-1)=0, so A/C= 0.
x=2: (2-1)(4)>0, so 1 is greatest positive # we can do. ok.
x=-1: (-2)(1) = -2. Now this is lowest #. Elim 0.
x=-2: (-3)(0) = 0. Not quite.
x=-3: (-4)(-1) = positive. Not going in right direction. **Answer must be -2<x<0.
x= -1/2: (-1/2-2/2)(-1/2+4/2)= (-3/2)(3/2)= -9/4. Good!
x= -3/2: (-3/2-2/2)(-3/2+4/2)= (-5/2)(1/2) = -5/4.
* -9/4 = -2.25; -5/4= -1.25.
-9/4 = correct. A/C B = correct.
Re: If y= (x-1)(x+2), then what is the least possible value of y   [#permalink] 05 Jul 2016, 11:52
Similar topics Replies Last post
Similar
Topics:
9 If Y = |X + 1| - |X-2|, then 7 16 Feb 2017, 14:13
10 What is least possible value of |23 - 7y| is (A)0 (B)1 (C)2 (D)5 (E)9 9 21 Feb 2017, 00:03
1 The Operation ==> is defined by x ==>y=x+(x+1)+(x+2)... +y. 1 20 Nov 2012, 03:11
52 If y is an integer, then the least possible value of |23-5| 17 07 Feb 2017, 09:38
12 If y is an integer, then the least possible value of |23–5| 4 05 Mar 2017, 09:56
Display posts from previous: Sort by