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the value of 1/ [√ (2 x) + √x]?

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by sanju09 » Fri Mar 23, 2012 3:12 am
If x > 0, then what is the value of 1/ [√ (2 x) + √x]?
A. 1/√ (3 x)
B. 1/ [2 √ (2 x)]
C. 1/ (x √2)
D. (√2 - 1)/ √x
E. (1 + √2)/ √x
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by mrinal85k » Fri Mar 23, 2012 3:24 am
Answer should be E[/list]
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by killer1387 » Fri Mar 23, 2012 3:55 am
sanju09 wrote:If x > 0, then what is the value of 1/ [√ (2 x) + √x]?
A. 1/√ (3 x)
B. 1/ [2 √ (2 x)]
C. 1/ (x √2)
D. (√2 - 1)/ √x
E. (1 + √2)/ √x
IMO D
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by ubhanja » Fri Mar 23, 2012 9:09 am
1/ [√ (2 x) + √x]= 1/√X(√2+1)=(√2-1)/√X(√2+1)(√2-1)=(√2-1)/√X

imo d
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by sanju09 » Sat Mar 24, 2012 1:31 am
ubhanja wrote:1/ [√ (2 x) + √x]= 1/√X(√2+1)=(√2-1)/√X(√2+1)(√2-1)=(√2-1)/√X

imo d
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by seal4913 » Sat Mar 24, 2012 11:44 am
ubhanja wrote:1/ [√ (2 x) + √x]= 1/√X(√2+1)=(√2-1)/√X(√2+1)(√2-1)=(√2-1)/√X

imo d
Please explain why/how you are getting (√2-1).
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by pemdas » Sat Mar 24, 2012 5:31 pm
seal4913 wrote:
ubhanja wrote:1/ [√ (2 x) + √x]= 1/√X(√2+1)=(√2-1)/√X(√2+1)(√2-1)=(√2-1)/√X

imo d
Please explain why/how you are getting (√2-1).
there's minor issue, and selected expression must be read as (√2-1)(√2+1)/√X(√2+1)
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