can someone help me please? please!!?

Question

what kind of maths is this? is this polynomials?

if x + y = 5, xy = 1 and x > y, then (square root of x + square root of y) / square root of x - square root of y) = ?

Answer 1

Suppose (√x + √y )/ (√x-√y) = a

Then (√x + √y ) = a (√x-√y)

Square both sides...
x+y+2 (√x √y) = a^2 (x+y-2(√x √y))

Now we know x+y=5 and xy=1 so
5 +2 (+/- 1) = a^2 (5-2 (+/-1))

So a^2 is 7/3 or 7/7 or 3/7 or 3/3

Thus a^2 is 7/3 or 1 or 3/7.
So a is √(7/3) or +1 or -1 or √(3/7)
.

Answer 2

Since xy = 1, y = 1/x

x + 1/x = 5 then .... or x^2 - 5x + 1 = 0

The roots are x = [5 +/- sqrt(21)]/2, with y being the reciprocals.

Algebraically you have [sqrt(x) + sqrt(y)]/[sqrt(x) - sqrt(y)].
Multiply by unity in the form [sqrt(x) + sqrt(y)]/[sqrt(x) +sqrt(y)] to get:

([sqrt(x) + sqrt(y)]^2)/ [x - y] = 7/(x - y), based on the initial conditions...which is 7/(2x - 5). Using the roots (above), you have:

+/- 7/sqrt(21) = +/- sqrt(7/3) (where "+/-" means "plus or minus")

Answer 3

square root of 7/3.

Answer 4

x + y = 5
xy = 1
y = 1/x
x + 1/x = 5
x^2 + 1 = 5x
x^2 - 5x + 1 = 0
x = (5 ± √(25 - 4))/2
x = (5 ± √21)/2
x > y
x = (5 + √21)/2
y = - (√21)/2
x = ((5 + √21)/2)
y = (- (√21)/2)

(√x + √y) / (√x - √y) =
(5 + j√66.8258) / 9.582576 =
0.5217804 + j√6.973674

Answer 5

you can start by using x=1/y
substitute it in your first equation to get
(1/y) + y = 5
simplifying gives y^2 -5*y +1 = 0

solving for the roots, we have

x=4.791287847
y=0.208712152

(sqrt(x)+sqrt(y))/(sqrt(x)-sqrt(y))=1.527525232

hope it helps

Answer 6

thats kinda hard, ? math is easy once you know how to work the formula, thats why teachers make sure you do alot of the same types of problems so you get down the formula, im sure you knew this and this is usless information but oh well, good luck, ask your teacher? some one on here will know it

Answer 7

These are polynomial equations, which is an equation in which a polynomial is set equal to another polynomial.

Answer 8

its (7/3)^(1/2)

Answer 9

( sq.rt. x + sq.rt. y )^2 = x + y + 2 sq.rt. (xy)
= 5 + 2 = 7
Therefore, sq.rt. x + sq.rt. y = sq.rt. 7 ... ... ( 1 )

Similarly, sq.rt. x - sq.rt. y = sq.rt. 3 ... ... ( 2 )

Therefore, (sq.rt. x + sq.rt. y) / (sq.rt. x - sq.rt. y) = sq.rt. (7/3)

Answer 10

given
x+y=5
xy = 1
x > y
find
[sqrt(x)+sqrt(y)]/[sqrt(x)-sqrt(y)]
solution
multiply nominator and denominator by
[sqrt(x)-sqrt(y)] and simplify
(x-y)/(x-2sqrt(x*y)+y)=
(x-y)/(5-2)=(x-y)/3