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# If x > 0, What is the least possible value of -2√(5x) + x + 9 ?

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Senior Manager
Joined: 18 Feb 2019
Posts: 482
Location: India
GMAT 1: 460 Q42 V13
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If x > 0, What is the least possible value of -2√(5x) + x + 9 ?  [#permalink]

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14 Apr 2019, 07:13
5
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45% (medium)

Question Stats:

61% (02:07) correct 39% (01:55) wrong based on 69 sessions

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If x > 0, What is the least possible value of -2√(5x) + x + 9 ?

A. 0
B. 1
C. √5
D. 4
E. 9
Director
Joined: 18 Jul 2018
Posts: 949
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
If x > 0, What is the least possible value of -2√(5x) + x + 9 ?  [#permalink]

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14 Apr 2019, 10:41
1
x > 0.

Least value of $$-2\sqrt{5x}+x+9$$
for least value, we want to try to cancel 9. if we consider x as 5.
Then $$-2\sqrt{5x}+x+9$$ = $$-2\sqrt{25}+5+9$$ = -10+5+9 = 4.

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Joined: 01 Oct 2018
Posts: 14
If x > 0, What is the least possible value of -2√(5x) + x + 9 ?  [#permalink]

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Updated on: 26 May 2019, 18:08
kiran120680 wrote:
If x > 0, What is the least possible value of -2√(5x) + x + 9 ?

A. 0
B. 1
C. √5
D. 4
E. 9

1 variant: (For this variant you need to know approx. root values of 2,3,5)
√5 ~ 2,25
√3 ~ 1,7
√2 ~ 1,4
so -4,5 * √x + x + 9
next let's test 1,4 and 9 , coz this numbers have nice roots
1: 6,5
4: 4
9: 5,5
Ok, let's test next 2 numbers: 5 and 6
5: -2* √(5*5) + 5 + 9 = 4
6: -4,5*√2*√3 + 6 + 9 = -4,5*1,4*1,7 + 6 + 9 = -4,3

2 variant:
f(x) = -2√(5x) + x + 9
f'(x) = -√(5/x) + 1
In oder to find maximum need
-√(5/x) + 1 = 0
x = 5
Input x = 5 in the equation
f(x)max = -2√(5*5) + 5 + 9 = 4

Originally posted by ArtyomYom on 26 May 2019, 17:09.
Last edited by ArtyomYom on 26 May 2019, 18:08, edited 1 time in total.
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Re: If x > 0, What is the least possible value of -2√(5x) + x + 9 ?  [#permalink]

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26 May 2019, 17:31
kiran120680 wrote:
If x > 0, What is the least possible value of -2√(5x) + x + 9 ?

A. 0
B. 1
C. √5
D. 4
E. 9

VeritasKarishma could you kindly explain this solution. Thank you.
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Re: If x > 0, What is the least possible value of -2√(5x) + x + 9 ?   [#permalink] 26 May 2019, 17:31
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