Note that:
(x - y)^2 = x^2 - 2xy + y^2
= (x^2 + 2xy + y^2) - 4xy
= (x + y)^2 - 4xy
= 5^2 - 4(1), since x + y = 5 and xy = 1
= 1.
Since x > y ==> x - y > 0, taking the positive square root of both sides gives:
x - y = 1.
By rationalizing the denominator:
(√x + √y)/(√x - √y)
= (√x + √y)^2/[(√x + √y)(√x - √y)]
= [x + 2√(xy) + y]/(x - y)
= [(x + y) + 2√(xy)]/(x - y)
= (5 + 2)/1, since x + y = 5, xy = 1, and x - y = 1
= 7.
I hope this helps!
solve system:
x+y=5
xy=1 -> y=1/x
----------------
x+1/x=5 /x
x^2-5x+1=0
x(1)=(5+√21)/2 , y(1)=(5-√21)/2
x(2)=(5-√21)/2 , y(2)=(5+√21)/2
X(1) > Y(1)
Rationalize denominator:
√x+√y/√x-√y=(√x+√y)^2/(x-y)
=(x+2√xy+y)/(x-y)
=( (x+y) + 2√xy) / (x-y)
=( 5+2)/(x-y)
=7/(x-y)
Now with x(1) and y(1) solve x-y like this:
x-y=(5+√21)/2 - (5-√21)/2
x-y=(5+√21- 5+√21) )/2
x-y=2√21/2
x-y=√21
Therefore:
√x+√y/√x-√y=7/√21
hope this helps you :)
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Answers & Comments
Verified answer
Note that:
(x - y)^2 = x^2 - 2xy + y^2
= (x^2 + 2xy + y^2) - 4xy
= (x + y)^2 - 4xy
= 5^2 - 4(1), since x + y = 5 and xy = 1
= 1.
Since x > y ==> x - y > 0, taking the positive square root of both sides gives:
x - y = 1.
By rationalizing the denominator:
(√x + √y)/(√x - √y)
= (√x + √y)^2/[(√x + √y)(√x - √y)]
= [x + 2√(xy) + y]/(x - y)
= [(x + y) + 2√(xy)]/(x - y)
= (5 + 2)/1, since x + y = 5, xy = 1, and x - y = 1
= 7.
I hope this helps!
solve system:
x+y=5
xy=1 -> y=1/x
----------------
x+1/x=5 /x
x^2-5x+1=0
x(1)=(5+√21)/2 , y(1)=(5-√21)/2
x(2)=(5-√21)/2 , y(2)=(5+√21)/2
X(1) > Y(1)
Rationalize denominator:
√x+√y/√x-√y=(√x+√y)^2/(x-y)
=(x+2√xy+y)/(x-y)
=( (x+y) + 2√xy) / (x-y)
=( 5+2)/(x-y)
=7/(x-y)
Now with x(1) and y(1) solve x-y like this:
x-y=(5+√21)/2 - (5-√21)/2
x-y=(5+√21- 5+√21) )/2
x-y=2√21/2
x-y=√21
Therefore:
√x+√y/√x-√y=7/√21
hope this helps you :)