Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
09 Jan 2021, 07:24
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
57% (02:43) correct
43%
(03:01)
wrong
based on 260
sessions
History
Date
Time
Result
Not Attempted Yet
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
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
09 Jan 2021, 08:19
Kudos
Bookmarks
Bunuel
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
09 Jan 2021, 08:31
Kudos
Bookmarks
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
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
