NEW HERE? LEARN MORE
closeclose
Last visit was: 25 Apr 2024, 04:59 It is currently 25 Apr 2024, 04:59
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92912
Own Kudos [?]: 618923 [11]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   09 Jan 2021, 07:24
11
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

65% (hard)

Question Stats:

63% (02:36) correct 37% (02:56) wrong based on 94 sessions
Hide Show timer Statistics
If the two roots \(r_1\) and \(r_2\) of the quadratic equation \(5x^2 - px + 1 = 0\) are such that \(|r_1 - r_2| = 1\), which of the following could be the value of \(p\)?


A. \(-3\sqrt 6\)

B. \(-2\sqrt 3\)

C. \(\sqrt 5\)

D. \(2\sqrt 3\)

E. \(3\sqrt 5\)


Are You Up For the Challenge: 700 Level Questions
ShowHide Answer
Official Answer
E
Most Helpful Reply
User avatar
L
chetan2u
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11173
Own Kudos [?]: 31909 [8]
Given Kudos: 290
Send PM
Reviews Badge
Re: If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   09 Jan 2021, 08:19
5
Kudos
3
Bookmarks
Expert Reply
Bunuel wrote:
If the two roots \(r_1\) and \(r_2\) of the quadratic equation \(5x^2 - px + 1 = 0\) are such that \(|r_1 - r_2| = 1\), what is the value of \(p\)?


A. \(-3\sqrt 5\)

B. \(-2\sqrt 3\)

C. \(\sqrt 5\)

D. \(2\sqrt 3\)

E. \(3\sqrt 5\)





If the quadratic equation is \(ax^2+bx+c=0\),
Sum of roots = \(\frac{-b}{a}\)
Product of roots = \(\frac{c}{a}\)


\(5x^2 -px + 1 = 0\) is the quadratic equation here.
Sum of roots = \(r_1+r_2=\frac{-(-p)}{5}=\frac{p}{5}\).....(i)
Product of roots = \(r_1r_2=\frac{1}{5}\)....(ii)

\(|r_1 - r_2| = 1\)
Square both sides.......
\((|r_1 - r_2|)^2 = 1^2\)..........
\(r_1^2+r_2^2-2r_1r_2=1\)............
\(r_1^2+r_2^2+2r_1r_2-4r_1r_2=1\).........
\((r_1+r_2)^2-4r_1r_2=1\)

Substitute the values of \(r_1+r_2 \ \ and \ \ r_1r_2\) from i and ii

\((r_1+r_2)^2-4r_1r_2=1\)............
\((\frac{p}{5})^2-4*\frac{1}{5}=1\)...............
\(\frac{p^2}{25}\)\(=1+\frac{4}{5}=\frac{9}{5}\)
\(p^2=25*\frac{9}{5}=5*9......p=3\sqrt{5}\)

E
General Discussion
User avatar
L
CrackverbalGMAT
GMAT Club Legend
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4946
Own Kudos [?]: 7626 [0]
Given Kudos: 215
Location: India
Send PM
Re: If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   09 Jan 2021, 08:31
Top Contributor
Dividing the equation by 5, we get \(x^2 - \frac{p}{5}x + \frac{1}{5} = 0\)

In the quadratic equation \(ax^2 + bx + c = 0\), the sum of roots = \(-\frac{b}{a}\) and the product of the roots = \(\frac{c}{a}\)

Therefore in the above equation, with roots \(r_1\) and \(r_2\)

\(r_1\) + \(r_2\) = \(\frac{p}{5}\) and \(r_1*r_2 = \frac{1}{5}\)

Also |\(r_1 - r_2 = 1\)|, so \(r_1 - r_2 = 1 \space -1\)

Using the algebraic expression \((a - b)^2 = (a + b)^2 - 4ab\)

\((r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1*r_2\)

\(1^2 = (\frac{p}{5})^2 - 4 * \frac{1}{5}\)

\(1 + \frac{4}{5} = \frac{p^2}{25}\)

\(\frac{9}{5} = \frac{p^2}{25}\)

\(9 = \frac{p^2}{5}\)

\(p^2 = 45\)

p = \(3\sqrt{5}\)


Option E

Arun Kumar
avatar
B
flash007
Intern
Intern
Joined: 28 Oct 2020
Posts: 15
Own Kudos [?]: 13 [0]
Given Kudos: 26
Send PM
Re: If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   09 Jan 2021, 09:01
CrackVerbalGMAT wrote:
Dividing the equation by 5, we get \(x^2 - \frac{p}{5}x + \frac{1}{5} = 0\)

In the quadratic equation \(ax^2 + bx + c = 0\), the sum of roots = \(-\frac{b}{a}\) and the product of the roots = \(\frac{c}{a}\)

Therefore in the above equation, with roots \(r_1\) and \(r_2\)

\(r_1\) + \(r_2\) = \(\frac{p}{5}\) and \(r_1*r_2 = \frac{1}{5}\)

Also |\(r_1 - r_2 = 1\)|, so \(r_1 - r_2 = 1 \space -1\)

Using the algebraic expression \((a - b)^2 = (a + b)^2 - 4ab\)

\((r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1*r_2\)

\(1^2 = (\frac{p}{5})^2 - 4 * \frac{1}{5}\)

\(1 + \frac{4}{5} = \frac{p^2}{25}\)

\(\frac{9}{5} = \frac{p^2}{25}\)

\(9 = \frac{p^2}{5}\)

\(p^2 = 45\)

p = \(3\sqrt{5}\)


Option E

Arun Kumar


Can you please explain why \(-3\sqrt{5}\) is not a possibility?
User avatar
L
Kinshook
GMAT Club Legend
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5343
Own Kudos [?]: 3964 [0]
Given Kudos: 160
Location: India
Schools: HBS '22 CBS '22 Wharton '22
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Send PM
Reviews Badge
Re: If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   10 Jan 2021, 10:13
Bunuel

p may be +ve or -ve
Both A & E are possible answers.
Please correct if needed.

chetan2u wrote:
Bunuel wrote:
If the two roots \(r_1\) and \(r_2\) of the quadratic equation \(5x^2 - px + 1 = 0\) are such that \(|r_1 - r_2| = 1\), what is the value of \(p\)?


A. \(-3\sqrt 5\)

B. \(-2\sqrt 3\)

C. \(\sqrt 5\)

D. \(2\sqrt 3\)

E. \(3\sqrt 5\)





If the quadratic equation is \(ax^2+bx+c=0\),
Sum of roots = \(\frac{-b}{a}\)
Product of roots = \(\frac{c}{a}\)


\(5x^2 -px + 1 = 0\) is the quadratic equation here.
Sum of roots = \(r_1+r_2=\frac{-(-p)}{5}=\frac{p}{5}\).....(i)
Product of roots = \(r_1r_2=\frac{1}{5}\)....(ii)

\(|r_1 - r_2| = 1\)
Square both sides.......
\((|r_1 - r_2|)^2 = 1^2\)..........
\(r_1^2+r_2^2-2r_1r_2=1\)............
\(r_1^2+r_2^2+2r_1r_2-4r_1r_2=1\).........
\((r_1+r_2)^2-4r_1r_2=1\)

Substitute the values of \(r_1+r_2 \ \ and \ \ r_1r_2\) from i and ii

\((r_1+r_2)^2-4r_1r_2=1\)............
\((\frac{p}{5})^2-4*\frac{1}{5}=1\)...............
\(\frac{p^2}{25}\)\(=1+\frac{4}{5}=\frac{9}{5}\)
\(p^2=25*\frac{9}{5}=5*9......p=3\sqrt{5}\)

E


Posted from my mobile device
User avatar
L
chetan2u
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11173
Own Kudos [?]: 31909 [0]
Given Kudos: 290
Send PM
Reviews Badge
Re: If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   10 Jan 2021, 10:24
Expert Reply
Kinshook wrote:
Bunuel

p may be +ve or -ve
Both A & E are possible answers.
Please correct if needed.

chetan2u wrote:
Bunuel wrote:
If the two roots \(r_1\) and \(r_2\) of the quadratic equation \(5x^2 - px + 1 = 0\) are such that \(|r_1 - r_2| = 1\), what is the value of \(p\)?


A. \(-3\sqrt 5\)

B. \(-2\sqrt 3\)

C. \(\sqrt 5\)

D. \(2\sqrt 3\)

E. \(3\sqrt 5\)





If the quadratic equation is \(ax^2+bx+c=0\),
Sum of roots = \(\frac{-b}{a}\)
Product of roots = \(\frac{c}{a}\)


\(5x^2 -px + 1 = 0\) is the quadratic equation here.
Sum of roots = \(r_1+r_2=\frac{-(-p)}{5}=\frac{p}{5}\).....(i)
Product of roots = \(r_1r_2=\frac{1}{5}\)....(ii)

\(|r_1 - r_2| = 1\)
Square both sides.......
\((|r_1 - r_2|)^2 = 1^2\)..........
\(r_1^2+r_2^2-2r_1r_2=1\)............
\(r_1^2+r_2^2+2r_1r_2-4r_1r_2=1\).........
\((r_1+r_2)^2-4r_1r_2=1\)

Substitute the values of \(r_1+r_2 \ \ and \ \ r_1r_2\) from i and ii

\((r_1+r_2)^2-4r_1r_2=1\)............
\((\frac{p}{5})^2-4*\frac{1}{5}=1\)...............
\(\frac{p^2}{25}\)\(=1+\frac{4}{5}=\frac{9}{5}\)
\(p^2=25*\frac{9}{5}=5*9......p=3\sqrt{5}\)

E


Posted from my mobile device


Kinshook you are correct. We could have the answer as A too. But now A has been changed
avatar
B
DanyP
Intern
Intern
Joined: 04 Mar 2019
Posts: 2
Own Kudos [?]: 0 [0]
Given Kudos: 2
Send PM
If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   18 Jan 2021, 18:05
isn't the algebraic identity (x-y)^2 = x^2 + y^2 -2xy?

Why is it that in the answers they are using, they use (x-y)^2 = x^2 +y^2 -4xy?
User avatar
L
CrackverbalGMAT
GMAT Club Legend
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4946
Own Kudos [?]: 7626 [1]
Given Kudos: 215
Location: India
Send PM
Re: If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   18 Jan 2021, 18:51
1
Kudos
Top Contributor
DanyP wrote:
isn't the algebraic identity (x-y)^2 = x^2 + y^2 -2xy?

Why is it that in the answers they are using, they use (x-y)^2 = x^2 +y^2 -4xy?



Hi DanyP. But I've written \((x - y)^2 = (x + y)^2 - 4xy\) not \(x^2 + y^2 - 4xy\)


Arun Kumar
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14822
Own Kudos [?]: 64908 [2]
Given Kudos: 426
Location: Pune, India
Send PM
Re: If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   19 Jan 2021, 03:57
1
Kudos
1
Bookmarks
Expert Reply
Bunuel wrote:
If the two roots \(r_1\) and \(r_2\) of the quadratic equation \(5x^2 - px + 1 = 0\) are such that \(|r_1 - r_2| = 1\), which of the following could be the value of \(p\)?


A. \(-3\sqrt 6\)

B. \(-2\sqrt 3\)

C. \(\sqrt 5\)

D. \(2\sqrt 3\)

E. \(3\sqrt 5\)


Are You Up For the Challenge: 700 Level Questions


Alternatively, you can try using the quadratic roots formula:

Roots = \(\frac{-b + \sqrt{b^2 - 4ac}}{2a}, \frac{-b - \sqrt{b^2 - 4ac}}{2a} \)

The difference between the roots is 1 which means the term \(\frac{\sqrt{b^2 - 4ac}}{2a}\) is 1/2. So one root is 1/2 more than -b/2a and the other is 1/2 less than -b/2a.

\(\frac{\sqrt{b^2 - 4ac}}{2a} = \frac{1}{2}\)

\(\frac{\sqrt{p^2 - 4*5}}{2*5} = \frac{1}{2}\)

\(p = +-3\sqrt{5}\)

Answer (E)
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32667
Own Kudos [?]: 821 [0]
Given Kudos: 0
Send PM
Re: If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink] New post   26 Dec 2023, 22:20
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.
GMAT Club Bot
gmatclubot
Re: If the two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0 [#permalink]
26 Dec 2023, 22:20
Moderators:
Bunuel
Math Expert
92912 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions