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 two distinct roots. How many integer values can k take?

A. 41
B. 40
C. 21
D. 20
E. 10

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yashikaaggarwal · Accepted Answer

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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chetan2u · Answer

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

A

firas92 · Answer

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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unraveled · Answer

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.

ArunSharma12 · Answer

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

MayankSingh · Answer

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

exc4libur · Answer

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

NitishJain · Answer

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

krosode20 · Answer

Should it not be 9 possible integer values. 20, 18, 14, 8, 0, -8, -14, -18, -20 are the possible values of K.

naveenban · Answer

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

HoneyLemon · Answer

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 .

Shobhit7 · Answer

|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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madgmat2019 · Answer

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

somenathd · Answer

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

sambitspm · Answer

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

monikakumar · Answer

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

Kinshook · Answer

Asked: 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?

For two distinct roots:-

81 - 4|k| > 0
|k| < 81/4 = 20.25
k = -20,-19,,,,0,1,2,.....19,20: 41 integers values

IMO A

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