Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 65187

If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 01:06
Question Stats:
50% (01:34) correct 50% (01:58) wrong based on 54 sessions
HideShow timer Statistics
Competition Mode Question If \(x^2  9x + k = 0\), where x is a variable and k is a constant, has two distinct roots. How many integer values can k take? A. 41 B. 40 C. 21 D. 20 E. 10 Are You Up For the Challenge: 700 Level Questions
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Math Expert
Joined: 02 Aug 2009
Posts: 8756

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
30 May 2020, 21:51
If \(x^2  9x + k = 0\), where x is a variable and k is a constant, has two distinct roots. How many integer values can k take? A. 41 B. 40 C. 21 D. 20 E. 10 krosode20, Why the answer will not be just the values you have taken is that x can be any value and NOT just integer. Everyone has been using Discriminant, which is not that widely used in GMAT, so let us do it by number properties. \(x^2  9x + k = 0........x(x9)=k\) So \(x(x9)\leq{0}\), that is \(0\leq{x}\leq{9}\) Now if it was given that x is integer, x would be 0, 1, 2..9 and possible values 0(09)=0..k=0 1(19)=8=8=8, so 8 and 8 2(29)=14=14=14, so 14 and 14 3(39)=18=18=18, so 18 and 18 4(49)=20=20=20, so 20 and 20 Total 9 values..But x can be anything and if x=3.2, then there will always be some real number y, such that 3.2*y=19 or 3.2*y=18 and so on. Therefore, we will be concerned with the maximum and minimum value that x can take.Minimum value :will be 0 Maximum value:We know if roots are a+b, then a+b=9 , and the product of two numbers, ab or k, is maximum when they are closest to each other, so here it will be a=b=9/2=4.5 so \(k=4.5*4.5=20.25\) \(20.25\leq{k}\leq{20.25}\) So integer values are \(20\leq{k}\leq{20}\)... Total = \(20(20)+1=41\) A
_________________




PS Forum Moderator
Joined: 18 Jan 2020
Posts: 1134
Location: India
GPA: 4

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 01:33
For two distinct real roots D>0 => D= b^24ac => B= 9, A = 1 and C=K => D = (9)^24*1*k > 0 => D = 814k>0 Let 814k=0 => 81=4k => k=20.25 => therefore, k has to be less than 20.25 => So K can take values from 1 to 20 = 20 => Since K is in module so ± values of K will always give + values. => for ex: k = 19 = 19 => so K will have values lying in set (20,19....1)(1,2,3.....20) = 40 values. => But putting K as zero will also give D>0 => So K can have 41 values in total. IMO A
Posted from my mobile device



Director
Joined: 16 Jan 2019
Posts: 659
Location: India
Concentration: General Management
WE: Sales (Other)

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 01:55
If \(x^29x+k=0\) has two distinct roots then \(9^24k(1)>0\)
So \(4k<81\) or \(k<20.25\)
Since \(k\) is an integer, \(k≤20\) or \(20≤k≤20\) which is \(41\) values
Answer is (A)
Posted from my mobile device



Sloan MIT School Moderator
Joined: 07 Mar 2019
Posts: 1296
Location: India
WE: Sales (Energy and Utilities)

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
Updated on: 29 May 2020, 02:21
If \(xˆ2−9x+k=0\), where x is a variable and k is a constant, has two distinct roots. How many integer values can k take? A. 41 B. 40 C. 21 D. 20 E. 10 Here k ≤ k ≤ +k Root of '\(xˆ2−9x+k=0\)' = \(\frac{9 ± \sqrt{814k}}{2}\) So, roots will depend on \(\sqrt{814k}\). Thus, \(\sqrt{814k}\) ≥ 0 Solving 81  4k = 0 \(k_{max} = 20\) and \(k_{min} = 0\) Therefore 20 ≤ k ≤ 20 i.e. k has 41 integer values. Answer A.
_________________
Originally posted by unraveled on 29 May 2020, 02:13.
Last edited by unraveled on 29 May 2020, 02:21, edited 1 time in total.



Director
Joined: 24 Oct 2015
Posts: 507
Location: India
GMAT 1: 650 Q48 V31
GPA: 4

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 02:18
If x2−9x+k=0, where x is a variable and k is a constant, has two distinct roots. How many integer values can k take?
A. 41 B. 40 C. 21 D. 20 E. 10
real and distinct roots: \(b^2 4ac > 0\) \(81 4k > 0\) \(4k < 81\) \(k < 20.25\) \(20.25 < k < 20.25\) number of k's is 41 Ans: A



Manager
Joined: 08 Jan 2018
Posts: 219
Location: India
Concentration: Operations, General Management
WE: Project Management (Manufacturing)

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 03:48
IMO A
x2−9x+k=0 For two distinct roots, D > 0 (a>0) b^24ac >0 9^2 4*1*k >0 k < 81/9 = 20.25 k <=20 20 <= K <=20 Total Numbers = 41
Ans . 41



SVP
Joined: 24 Nov 2016
Posts: 1674
Location: United States

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 04:59
Quote: If x2−9x+k=0, where x is a variable and k is a constant, has two distinct roots. How many integer values can k take?
A. 41 B. 40 C. 21 D. 20 E. 10 x^2−9x+k=0 has two dif roots discriminant b^24ac>0 (9)^24(1)(k)>0 4k<81, 20.25<k<20.25 integers bet: 20(20)+1=41 Ans (A)



IESE School Moderator
Joined: 11 Feb 2019
Posts: 308

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 05:29
A. Given: x^2  9x + k = 0 b^24ac >=0 814k >=0 20.25>=k 20.25 =< k =< 20.25 ==> 41 integer values of k
_________________



Intern
Joined: 06 Sep 2018
Posts: 2

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 07:53
Should it not be 9 possible integer values. 20, 18, 14, 8, 0, 8, 14, 18, 20 are the possible values of K.



Manager
Joined: 10 Oct 2017
Posts: 81

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 08:45
Discriminant, D = 81 – 4k If roots are real, D > 0 81 – 4k > 0 4k < 81 k < 20.25
Hence, –20.25 < k < 20.25 The integer values that k can take are –20, –19, –18 … 0 … 18, 19 and 20.
41 different values
Option A



Stern School Moderator
Joined: 26 May 2020
Posts: 271
Concentration: General Management, Technology
WE: Analyst (Computer Software)

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
29 May 2020, 08:54
A, IMO. If x^2−9x+k=0 , where x is a variable and k is a constant, has two distinct roots. so 9^2  4 k >0 81/4 > K 20.25 > K so 20 <= K <= 20 so K can have 20 + 20 + 1 (for 0) = 41 integer values . So A .
_________________
Thank you. Regards, Ashish A Das.
The more realistic you are during your practice, the more confident you will be during the CAT.



Senior Manager
Joined: 01 Feb 2017
Posts: 271

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
30 May 2020, 12:22
k<81/4 k<20.25
Min k=0 Max k=20 Abs k can take 21 values
So, possible values of k: 20+ve, 20ve, and Zero = 41
Ans A
Posted from my mobile device



Director
Joined: 01 Mar 2019
Posts: 572
Location: India
Concentration: Strategy, Social Entrepreneurship
GPA: 4

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
30 May 2020, 20:51
As they are distinct roots
D>0 b^24ac>0 814k>0 4k<81 k< 20.25
20.25 < k < 20.25
total 41 values including 0
OA:A



Intern
Joined: 11 Nov 2018
Posts: 2

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
30 May 2020, 21:07
b^24ac>=0 814 * k >=0  k =<20.25 20.25<k<20.25
Total number integers within given range should be 41.
Posted from my mobile device



Senior Manager
Joined: 05 Aug 2019
Posts: 301
Location: India
Concentration: Leadership, Technology
GMAT 1: 600 Q50 V22
GPA: 4

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
31 May 2020, 21:26
x^29x+k = 0 2 distinct roots
Determinant has to be >0 b^24ac > 0
81  4k > 0 k< 81/4
20<=k<= 20 Hence number is 20+20+ 1 = 41 Hence A



Manager
Joined: 23 Jan 2020
Posts: 171

Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
Show Tags
31 May 2020, 21:39
If x2−9x+k=0, where x is a variable and k is a constant, has two distinct roots. How many integer values can k take?
A. 41 B. 40 C. 21 D. 20 E. 10
2 distinct roots b^24ac >0 814k>0 20.25<k<20.25 k can be 20+20+1 41 Ans A




Re: If x^2  9x + k = 0, where x is a variable and k is a constant, has
[#permalink]
31 May 2020, 21:39




