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If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has

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If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 29 May 2020, 01:06
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 30 May 2020, 21:51
1
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.
\(x^2 - 9x + |k| = 0........x(x-9)=-|k|\)
So \(x(x-9)\leq{0}\), 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(0-9)=0..k=0
1(1-9)=-8=-|8|=-|-8|, so 8 and -8
2(2-9)=-14=-|14|=-|-14|, so 14 and -14
3(3-9)=-18=-|18|=-|-18|, so 18 and -18
4(4-9)=-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 = \(20-(-20)+1=41\)

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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 29 May 2020, 01:33
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2
For two distinct real roots D>0
=> D= b^2-4ac
=> B= -9, A = 1 and C=K
=> D = (-9)^2-4*1*k > 0
=> D = 81-4k>0
Let 81-4k=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

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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 29 May 2020, 01:55
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If \(x^2-9x+|k|=0\) has two distinct roots then \(9^2-4|k|(1)>0\)

So \(4|k|<81\) or \(|k|<20.25\)

Since \(k\) is an integer, \(|k|≤20\) or \(-20≤k≤20\) which is \(41\) values

Answer is (A)

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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post Updated on: 29 May 2020, 02:21
1
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 '\(xˆ2−9x+|k|=0\)' = \(\frac{9 ± \sqrt{81-4|k|}}{2}\)
So, roots will depend on \(\sqrt{81-4|k|}\).

Thus, \(\sqrt{81-4|k|}\) ≥ 0
Solving 81 - 4|k| = 0
\(|k|_{max} = 20\) and \(|k|_{min} = 0\)

Therefore -20 ≤ |k| ≤ 20 i.e. k has 41 integer values.

Answer A.
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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.
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 29 May 2020, 02:18
1
1
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: \(b^2 -4ac > 0\)
\(81 -4|k| > 0\)
\(4|k| < 81\)
\(|k| < 20.25\)
\(-20.25 < k < 20.25\)
number of k's is 41
Ans: A
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 29 May 2020, 03:48
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IMO A

x2−9x+|k|=0
For two distinct roots, D > 0 (a>0)
b^2-4ac >0
9^2- 4*1*|k| >0
|k| < 81/9 = 20.25
|k| <=20
-20 <= K <=20
Total Numbers = 41

Ans . 41
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 29 May 2020, 04:59
1
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^2-4ac>0
(-9)^2-4(1)(|k|)>0
4|k|<81, -20.25<k<20.25
integers bet: 20-(-20)+1=41

Ans (A)
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 29 May 2020, 05:29
1
A.
Given: x^2 - 9x + |k| = 0

b^2-4ac >=0
81-4|k| >=0
20.25>=|k|

-20.25 =< k =< 20.25

==> 41 integer values of k
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 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.
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 29 May 2020, 08:45
2
Discriminant, D = 81 – 4|k|
If roots are real, D > 0
81 – 4|k| > 0
4|k| < 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
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 29 May 2020, 08:54
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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 .
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 30 May 2020, 12:22
1
|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, 20-ve, and Zero = 41

Ans A

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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 30 May 2020, 20:51
1
As they are distinct roots

D>0
b^2-4ac>0
81-4|k|>0
4|k|<81
|k|< 20.25

-20.25 < k < 20.25

total 41 values including 0

OA:A
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 30 May 2020, 21:07
b^2-4ac>=0
81-4 *| k |>=0
| k |=<20.25
-20.25<k<20.25

Total number integers within given range should be 41.

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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 31 May 2020, 21:26
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x^2-9x+|k| = 0
2 distinct roots

Determinant has to be >0
b^2-4ac > 0

81 - 4|k| > 0
|k|< 81/4

-20<=k<= 20
Hence number is 20+20+ 1 = 41
Hence A
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has  [#permalink]

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New post 31 May 2020, 21:39
1
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^2-4ac >0
81-4|k|>0
-20.25<k<20.25
k can be 20+20+1
41
Ans A
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Re: If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has   [#permalink] 31 May 2020, 21:39

If x^2 - 9x + |k| = 0, where x is a variable and k is a constant, has

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