GRE Prep 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.
Director
Joined: 16 May 2014
Posts: 591
Given Kudos: 0
GRE 1: Q165 V161
If one of the roots of the equation
[#permalink]
27 Mar 2016, 23:13
27 Mar 2016, 23:13
Expert Reply
00:00
Question Stats:
43% (01:30) correct
56% (02:25) wrong based on 16 sessions
Hide Show timer Statistics
If one of the roots of the equation \(2x^2\) + (3k + 4)x + (\(9k^2\)–3k– 1) = 0 is twice the other, then which of the following can be a value of ‘k’?
A. \(\frac{-2}{3}\)
B. \(\frac{2}{3}\)
C. \(\frac{-1}{3}\)
D. \(\frac{1}{3}\)
E. None of the above
A. \(\frac{-2}{3}\)
B. \(\frac{2}{3}\)
C. \(\frac{-1}{3}\)
D. \(\frac{1}{3}\)
E. None of the above
Director
Joined: 16 May 2014
Posts: 591
Given Kudos: 0
GRE 1: Q165 V161
Re: If one of the roots of the equation
[#permalink]
28 Mar 2016, 00:30
28 Mar 2016, 00:30
2
Expert Reply
Explanation
Let one of the roots of the given equation be ‘a’. Then, the other root will be ‘2a’.
∴ Sum of the roots = a + 2a = -(3k+4)/2
or, a = -(3k+4)/6 ----- (i)
∴ Product of the roots = a*2a = \(2a^2\)= \((9k^2-3k-1)/2\)
or, \(a^2\) = \((9k^2-3k-1)/4\) ---- (ii)
Combining (i) and (ii),
⇒ \(72k^2\)−51k−25 = 0
⇒(3k + 1)(24k − 25) = 0
Therefore, k = -1/3, 25/24.
Re: If one of the roots of the equation
[#permalink]
04 Jun 2019, 01:02
04 Jun 2019, 01:02
I did the backward calculation by taking the possible values of K. I tried to find whether the reformed equation ( in line with the value of K) gives two roots or not and the roots corroborate to the conditions or not. Eventually , whiling taking K=-1/3, It matched the conditions. Thus, I got C.
But this was too time-consuming.
Can any of you please give an easy and quicker solution ?
Regards
But this was too time-consuming.
Can any of you please give an easy and quicker solution ?
Regards
