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805+ (Hard)|   Algebra|      
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gmihir
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists Updated on: 20 Feb 2016, 00:22
Originally posted by gmihir on 13 May 2012, 08:16.
Last edited by Bunuel on 20 Feb 2016, 00:22, edited 1 time in total.
Edited the question.
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D
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95% (hard)

Question Stats:

43% (02:44) correct 57% (03:01) wrong based on 114 sessions

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If \(x = -5 + \sqrt{(45 +4k – k^2 )}\) where k is a positive integer. How many values of k exists if x has to be positive?

A. 9
B. 8
C. 7
D. 6
E. 5
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D
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists 20 Feb 2016, 00:51
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gmihir
If \(x = -5 + \sqrt{(45 +4k – k^2 )}\) where k is a positive integer. How many values of k exists if x has to be positive?

A. 9
B. 8
C. 7
D. 6
E. 5
Show more

Hi,
if we take the Q in the present state..
\(x = -5 + \sqrt{(45 +4k – k^2 )}\) ..
since x has to be positive, the RHS \(-5 + \sqrt{(45 +4k – k^2 )}\) should also be positive..
\(-5 + \sqrt{(45 +4k – k^2 )}>0..\)..
\(\sqrt{(45 +4k – k^2 )}>5\).
\((45 +4k – k^2 )>25\)..
\(45 +4-4+4k-k^2>25\)..
\(-(k-2)^2>-24\)...
or \((k-2)^2<24\)...
the square of 4 is lessthan 24 and 5 is greater than 24..
so (k-2)<5..hence k<7
therefore k can take 6 values
D
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists 13 May 2012, 08:48
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I am getting a total of 8 solutions i.e B
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists 13 May 2012, 09:01
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For x = -5 + ( 45 +4k – k^2 )0.5 to be positive , 45 +4k – k^2 should be greater than 10.
45 +4k – k^2 > 10
multiplying the inequality by -ve.
k^2-4k -35<0
which gives k = (4+/-12.5 )/2
(k-8.25) (k+4.12) < 0

On the number lie k would lie between , -4.12 <k < 8.25 , since K is to be positive so k =1,2,3,4,5,6,7,8

Hence ,B .
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists 19 Feb 2016, 21:44
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I got B too. Can anyone help me out with the official solution? Thanks
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists 19 Feb 2016, 22:10
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Kingsman
I got B too. Can anyone help me out with the official solution? Thanks
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Hi,
I do not think the person who posted the Q is any more active, so official Solution may not be known..

But let me tell you a way, which should be close to being official solution..

Quote:
If x = -5 + ( 45 +4k – k^2 )0.5 where k is a positive integer. How many values of k exists if x has to be positive?

A. 9
B. 8
C. 7
D. 6
E. 5..
Show more

lets simplify the equation further..
x = -5 + ( 45 +4k – k^2 )0.5 ..
x=(-10+ 45 +4k – k^2 )/2..
x=(35 +4k – k^2 )/2..
x=(35+4-4 +4k – k^2 )/2..(add and subtract 4, so that we can further simplify)
x={39-( k-2)^2}/2..
now for x to be positive, {39-( k-2)^2} should be positive ..
so {39-( k-2)^2}>0..
39>( k-2)^2..
now we know taht the greatest positive integer, whose square is just lesser than 39 is 6(square =36)..
so k-2=6 or k=8 ans
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists 20 Feb 2016, 00:24
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chetan2u
Kingsman
I got B too. Can anyone help me out with the official solution? Thanks


Hi,
I do not think the person who posted the Q is any more active, so official Solution may not be known..

But let me tell you a way, which should be close to being official solution..

Quote:
If x = -5 + ( 45 +4k – k^2 )0.5 where k is a positive integer. How many values of k exists if x has to be positive?

A. 9
B. 8
C. 7
D. 6
E. 5..

lets simplify the equation further..
x = -5 + ( 45 +4k – k^2 )0.5 ..
x=(-10+ 45 +4k – k^2 )/2..
x=(35 +4k – k^2 )/2..
x=(35+4-4 +4k – k^2 )/2..(add and subtract 4, so that we can further simplify)
x={39-( k-2)^2}/2..
now for x to be positive, {39-( k-2)^2} should be positive ..
so {39-( k-2)^2}>0..
39>( k-2)^2..
now we know taht the greatest positive integer, whose square is just lesser than 39 is 6(square =36)..
so k-2=6 or k=8 ans
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Edited the question. It should have been: x = -5 + ( 45 +4k – k^2 )^0.5.
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists 20 Feb 2016, 06:56
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chetan2u
gmihir
If \(x = -5 + \sqrt{(45 +4k – k^2 )}\) where k is a positive integer. How many values of k exists if x has to be positive?

A. 9
B. 8
C. 7
D. 6
E. 5

Hi,
if we take the Q in the present state..
\(x = -5 + \sqrt{(45 +4k – k^2 )}\) ..
since x has to be positive, the RHS \(-5 + \sqrt{(45 +4k – k^2 )}\) should also be positive..
\(-5 + \sqrt{(45 +4k – k^2 )}>0..\)..
\(\sqrt{(45 +4k – k^2 )}>5\).
\((45 +4k – k^2 )>25\)..
\(45 +4-4+4k-k^2>25\)..
\(-(k-2)^2>-24\)...
or \((k-2)^2<24\)...
the square of 4 is lessthan 24 and 5 is greater than 24..
so (k-2)<5..hence k<7
therefore k can take 6 values
D
Show more

Thanks Chetan2u and Bunuel :)
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists 20 Feb 2016, 10:00
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chetan2u
gmihir
If \(x = -5 + \sqrt{(45 +4k – k^2 )}\) where k is a positive integer. How many values of k exists if x has to be positive?

A. 9
B. 8
C. 7
D. 6
E. 5

Hi,
if we take the Q in the present state..
\(x = -5 + \sqrt{(45 +4k – k^2 )}\) ..
since x has to be positive, the RHS \(-5 + \sqrt{(45 +4k – k^2 )}\) should also be positive..
\(-5 + \sqrt{(45 +4k – k^2 )}>0..\)..
\(\sqrt{(45 +4k – k^2 )}>5\).
\((45 +4k – k^2 )>25\)..
\(45 +4-4+4k-k^2>25\)..
\(-(k-2)^2>-24\)...
or \((k-2)^2<24\)...
the square of 4 is lessthan 24 and 5 is greater than 24..
so (k-2)<5..hence k<7
therefore k can take 6 values
D
Show more

thanks chetan2u for that realization.
I dont know how people got 8.
what are u rearranging?
k CANT be more than 6. if k = 7 then x will not be positive.
the square root side needs to be more than 25 to cancel out negative 25.
clear D
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x = -5 + ( 45 +4k – k^2 )0.5 How many values of k exists 20 Feb 2016, 10:11
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chetan2u
gmihir
If \(x = -5 + \sqrt{(45 +4k – k^2 )}\) where k is a positive integer. How many values of k exists if x has to be positive?

A. 9
B. 8
C. 7
D. 6
E. 5

Hi,
if we take the Q in the present state..
\(x = -5 + \sqrt{(45 +4k – k^2 )}\) ..
since x has to be positive, the RHS \(-5 + \sqrt{(45 +4k – k^2 )}\) should also be positive..
\(-5 + \sqrt{(45 +4k – k^2 )}>0..\)..
\(\sqrt{(45 +4k – k^2 )}>5\).
\((45 +4k – k^2 )>25\)..
\(45 +4-4+4k-k^2>25\)..
\(-(k-2)^2>-24\)...
or \((k-2)^2<24\)...
the square of 4 is lessthan 24 and 5 is greater than 24..
so (k-2)<5..hence k<7
therefore k can take 6 values
D

thanks chetan2u for that realization.
I dont know how people got 8.
what are u rearranging?
k CANT be more than 6. if k = 7 then x will not be positive.
the square root side needs to be more than 25 to cancel out negative 25.
clear D
Show more

People got 8 as the question poster has mentioned the question as \(x = -5 + (45 +4k – k^2 )0.5\) instead of \(x = -5 + \sqrt{(45 +4k – k^2 )}\)
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