GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 18 Aug 2018, 07:50

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Let p and q be the roots of the quadratic equation

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
B
Joined: 01 Apr 2018
Posts: 23
Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 09 Jun 2018, 06:35
4
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

51% (02:08) correct 49% (02:06) wrong based on 69 sessions

HideShow timer Statistics

Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12

_________________

Please give me KUDOS if my post helps you...

Senior Manager
Senior Manager
User avatar
G
Joined: 24 Aug 2016
Posts: 282
Location: India
Concentration: Entrepreneurship, Operations
GMAT 1: 540 Q49 V16
GMAT 2: 640 Q47 V31
GPA: 3.4
Reviews Badge CAT Tests
Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 09 Jun 2018, 07:10
tarunanandani wrote:
Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12




\(x^{2}-(a-2)x-a-1=0\) = \(x^{2}-(a-2)x+(-a-1)=0\) equivalent to \(x^{2}-(p+q)x+pq=0\)
Hence \(p+q=a-2\) & \(pq=-a-1\)
\(p^{2} + q^{2}\)= \((p+q)^2-2pq\) = \((a-2)^2-2(-a-1)\) = \(a^2-4a+4+2a+2\) = \(a^2-2a+6\)

* When a =0 or 2 , \(p^{2} + q^{2}\)=6 but this is not the min value.

Now , \(a^2-2a+6\) is min when \(a^2-2a\) is minimum i.e, a=1
Hence \(p^{2} + q^{2}\)= 6-1=5.......................................................Hence I would go for option D.
_________________

Please let me know if I am going in wrong direction.
Thanks in appreciation.

Director
Director
User avatar
G
Status: Learning stage
Joined: 01 Oct 2017
Posts: 505
WE: Supply Chain Management (Energy and Utilities)
Premium Member CAT Tests
Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 09 Jun 2018, 07:20
tarunanandani wrote:
Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12


Lets equate the co-efficient s of given quadratic equation with the standard form \(ax^2+bx+c=0\)

a=1,b=-(a-2),c=-(a+1)

Sum of roots, p+q=\(-\frac{b}{a}\)=-(-(a-2))=a-2

product of roots, pq=\(\frac{c}{a}\)=-(a+1)

Now, \(p^2+q^2\)=\((p+q)^2-2pq\)=\((a-2)^2+2(a+1)\)=\(a^2-2a+6\)=\((a-1)^2+5\)

Now, min(\(p^2+q^2\))=min(\((a-1)^2\))+5)=0+5=5 (Since minimum value of any square expression is zero.)

Ans. D
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3042
Location: India
GPA: 3.12
Premium Member CAT Tests
Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 09 Jun 2018, 07:47
tarunanandani wrote:
Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12


Quadratic equation \(x^{2} - (a-2)x -a-1 = 0\) can be re-written as \(x^{2} - (a-2)x -(a+1) = 0\)

For a quadratic equation \(ax^2 + bx + x\). The sum of the roots is \(\frac{-b}{a}\) and product of the roots is \(\frac{c}{a}\)

Sum of the roots(p+q) = \((a-2)\) | Product of the roots(pq) = \(-(a+1)\)

\(p^{2} + q^{2} = (p+q)^2 - 2*pq\)

Here, \(p^{2} + q^{2} = (a-2)^2 - 2(-a-1) = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6\)

Since we need the minimum value, a^2 - 2a = -1 when a = 1. Therefore, \(p^{2} + q^{2} = 5\) (Option C)
_________________

You've got what it takes, but it will take everything you've got

Intern
Intern
avatar
B
Joined: 10 Oct 2017
Posts: 3
Concentration: Marketing
GPA: 3.71
Re: Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 11 Jun 2018, 14:18
How do you figure out that p^2+q^2 is equal to (p+q)^2-2pq?
BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3042
Location: India
GPA: 3.12
Premium Member CAT Tests
Re: Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 11 Jun 2018, 14:27
1
dmod14 wrote:
How do you figure out that p^2+q^2 is equal to (p+q)^2-2pq?


Hi dmod14

You must be aware of the property: \((p+q)^2 = p^2 + 2pq + q^2\)

Re-arranging this we get \(p^2+q^2 = (p+q)^2 - 2pq\)

Hope this helps you!
_________________

You've got what it takes, but it will take everything you've got

Intern
Intern
avatar
B
Joined: 10 Oct 2017
Posts: 3
Concentration: Marketing
GPA: 3.71
Re: Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 11 Jun 2018, 15:05
Thanks pushpitkc! That makes a lot of sense.

Can you explain how you got the minimum value, specifically how you went from a^2-2a+6 to p^2+q^2=5
Manager
Manager
avatar
S
Joined: 04 Aug 2010
Posts: 235
Schools: Dartmouth College
Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 11 Jun 2018, 16:39
1
1
tarunanandani wrote:
Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12

For any quadratic in the form \(x^2 + bx + c=0\):
Product of the roots = c.
Sum of the roots = -b.

Converting \(x^{2}-(a-2)x-a-1=0\) into the form \(x^2 + bx + c=0\), we get:
\(x^{2} + (-a+2)x + (-a-1)=0\).

In the quadratic above, c = -a-1 and -b = a-2.
Since the roots are p and q, we get:
Product of the roots = pq = -a-1.
Sum of the roots = p+q = a-2.

Squaring the blue equation, we get:
\((p+q)^2\) = \((a-2)^2\)
\(p^2 + q^2 + 2pq\) = \(a^2 + 4 - 4a\)

Substituting pq = -a-1 into the red equation, we get:
\(p^2 + q^2 + 2(-a-1)\) = \(a^2 + 4 - 4a\)
\(p^2 + q^2 -2a - 2\) = \(a^2 + 4 - 4a\)
\(p^2 + q^2\) = \(a^2 - 2a + 6\)

To minimize the value of \(p^2 + q^2\), we must minimize the value of \(a^2 - 2a + 6\).
Test easy values in \(a^2 - 2a + 6\):
a = 0 --> \(a^2 - 2a + 6 = 6\)
a = 1 --> \(a^2 - 2a + 6 = 5\)
a = 2 --> \(a^2 - 2a + 6 = 6\)
The resulting values on the right -- 6, 5, 6 -- indicate that \(y = a^2 - 2a + 6\) is a U-shaped parabola with a minimum value of 5.
Thus:
The minimum value of \(p^2 + q^2 = 5\).


_________________

GMAT and GRE Tutor
Over 1800 followers
Click here to learn more
GMATGuruNY@gmail.com
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.

Senior Manager
Senior Manager
User avatar
G
Joined: 29 Dec 2017
Posts: 338
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
GPA: 3.25
WE: Marketing (Telecommunications)
CAT Tests
Re: Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 11 Jun 2018, 19:33
1
dmod14 wrote:
Thanks pushpitkc! That makes a lot of sense.

Can you explain how you got the minimum value, specifically how you went from a^2-2a+6 to p^2+q^2=5


a^2-2a+6 = p^2+q^2, and you are asked to find min of p^2+q^2, so you have to find the minimum of a^2-2a+6 - parabola
The minimum of parabola is point x= -b/2a=-(-2)/2*1=1, so y will be 1^2-2*1+6=5. Hence min p^2+q^2=5
_________________

I'm looking for a study buddy in NY, who is aiming at 700+. PM me.

BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3042
Location: India
GPA: 3.12
Premium Member CAT Tests
Re: Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 11 Jun 2018, 22:57
1
dmod14 wrote:
Thanks pushpitkc! That makes a lot of sense.

Can you explain how you got the minimum value, specifically how you went from a^2-2a+6 to p^2+q^2=5


Hey dmod14

If you observe the solution, you will arrive at \(p^2+q^2 = a^2-2a+6\)

We have been asked to find the minimum value of the expression \(p^2+q^2\).
This will happen when the expression \(a^2-2a\) takes the minimum value.
For that to happen, we test values and find that when a = 1 , \(a^2-2a\) = -1.
Now, the value of the overall expression \(a^2-2a +6\) now becomes \(-1+6 = 5\)

Hope this helps you!
_________________

You've got what it takes, but it will take everything you've got

Intern
Intern
avatar
B
Joined: 01 Apr 2018
Posts: 23
Re: Let p and q be the roots of the quadratic equation  [#permalink]

Show Tags

New post 12 Jun 2018, 04:32
1
dmod14 wrote:
Thanks pushpitkc! That makes a lot of sense.

Can you explain how you got the minimum value, specifically how you went from a^2-2a+6 to p^2+q^2=5


The equation \(p^{2}+q^{2}=a^{2}−2a+6\) can be re-written as \(p^{2}+q^{2}=(a^{2}−2a+1^{2})+5\).
Now the last equation can be simplified as \(p^{2}+q^{2}=(a-1)^{2}+5\) and for \(p^{2}+q^{2}\) to be minimum the term \((a-1)^{2}\) has to be minimum, i.e., 0 as square of any term can't be negative. Therefore the minimum value of \(p^{2}+q^{2}=0+5=5\).
_________________

Please give me KUDOS if my post helps you...

Re: Let p and q be the roots of the quadratic equation &nbs [#permalink] 12 Jun 2018, 04:32
Display posts from previous: Sort by

Let p and q be the roots of the quadratic equation

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.