Find the square root of a surd term

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  • #1
chwala
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Homework Statement:

find the square root of ## a+b+√(2ab +b^2)##

Relevant Equations:

square root
find the square root of ## a+b+√(2ab +b^2)##

let ##√[a+b+√(2ab +b^2)]= ±(√x +√y)##
then, ##a+b+√(2ab +b^2)= x+y+ 2√(xy)##
where ##a+b=x+y##....................................1
##(b+a)^2-a^2=4xy## .....................2
from 2,
##a^2=(b+a)^2-4xy##
##a=√[x+y)^2-4xy]##
##a=√[x^2-2xy+y^2]##
##a=x-y##
therefore,
##b=x+y-x+y##
##b=2y##
##y=\frac {b}{2}##
→##x=a+\frac {b}{2}##
this is my original working...i am looking for any alternative method?
 
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Answers and Replies

  • #2
anuttarasammyak
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After your good effort we can show it in "2 in front of square root" way as
[tex]a+b+2\sqrt{(a+\frac{b}{2})\frac{b}{2}}[/tex]
 
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  • #3
chwala
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After your good effort we can show it in "2 in front of square root" way as
[tex]a+b+2\sqrt{(a+\frac{b}{2})\frac{b}{2}}[/tex]
why are you going ahead to find the products of the square root? the question simply asked to find the square root and not product of the square root. The square root will be equal to
##±(√x +√y)= ±(√[(a+\frac {b}{2}]##+##√\frac {b}{2}##

looking at your expression, it gives me ## (√x +√y)(√x +√y)= x+y+2√(xy)##
which is not what we are looking for...
 
  • #4
chwala
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ok i see what you meant...it was in reference to the equation 2...bingo
 
  • #5
chwala
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1593401236240.png

This is a solution from a colleague...
 

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