Question

\[x^2-y^2=a,2xy=b,\text{find }x^2+y^2\]

Answer 1

in terms of a and b

Answer 2

subtract the second expression from the first one and see if you can spot something

Answer 3

\[x^2-(\frac{b}{2x})^2=x^2-\frac{b^2}{4x^2}=a\]
\[\text{let }x^2=k\]
\[4k^2-4ak-b^2=0\]
\[k=\frac{4a\pm \sqrt{16a^2+16b}}{8}=\frac{a\pm \sqrt{a^2+b}}{2}\]
so \[y=\frac{b^2}{4\frac{a\pm \sqrt{a^2+b}}{2}}=\frac{b^2}{2 a\pm \sqrt{a^2+b}}\]
\[x^2+y^2=\frac{a\pm \sqrt{a^2+b}}{2}+\frac{b^2}{2 a\pm \sqrt{a^2+b}}\]

Answer 4

\[x^2-y^2-2xy=(x-y)^2\]

Answer 5

yup - then factorise the first expression

Answer 6

wait
somethings wrong

Answer 7

\[x^2+y^2+2xy=(x+y)^2=a+b\]

Answer 8

where did you get that from?

Answer 9

oh its when i open this
\[(x+y)^2=x^2+2xy+y^2\]

Answer 10

I mean where did you get the right-hand-side from (a+b)?

Answer 11

ohh mistake

Answer 12

:)

Answer 13

this question is actually a follow up of this
http://openstudy.com/study#/updates/51fd18d1e4b05ca7841f512c

Answer 14

ok - but if you follow my suggestion then you should be able to solve this

Answer 15

\[x^2-y^2-2xy=?\]

Answer 16

a-b

Answer 17

i.e. you know:\[x^2-y^2-2xy=a-b\]therefore:\[(x-y)^2=a-b\]therefore:\[x-y=\sqrt{a-b}\tag{1}\]and from the first expression you get:\[x^2-y^2=a\]therefore:\[(x+y)(x-y)=a\]then use expression (1) to find x+y

Answer 18

then use:\[(x+y)^2=x^2+y^2+2xy\]

Answer 19

\[ (x+y)^2=\frac{a^2}{(\sqrt{a-b})^2}\]

so \[x^2+y^2=\frac{a^2}{a-b}-b\]

Answer 20

\[\frac{a^2+b^2-ab}{a-b}\]

Answer 21

looks right

Answer 22

if we go back to our original problem

\[(x^2+y^2)^2=\frac{(a^2+b^2-ab)^2}{(a-b)^2}\] is required

Answer 23

when you say "original problem" do you mean the one referred to by the link you posted above?

Answer 24

yes the one on the link,for case \[a=\frac{ac+bd}{c^2+d^2},b=\frac{bc+ad}{c^2+d^2}\]

Answer 25

let me take a look at the original to see if there is an easier way to get to the solution...

Answer 26

yes the one on the link,for case \[a=\frac{ac+bd}{c^2+d^2},b=\frac{bc+ad}{c^2+d^2}\]

Answer 27

\[b=\frac{bc-ad}{c^2+d^2}\]
we need to show that
\[(x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}\]

Answer 28

let x = sqrta cost
and y = i sqrta sint

so
2ia sint cost = b

ia sin2t = b

you have to find value of a(cos^2 t - sin^2 t) or acos2t
= asqrt(1- sin^2 (2t))

just substitute value of sin2t

Answer 29

\[\text{let x} = \sqrt a \cos t\]
and \[y = i \sqrt a \sin t\]

so
\[2ia \sin t \cos t = b\]

\[ia \sin2t = b\]

you have to find value of \[a(\cos^2 t -\sin^2 t) ,OR a\cos2t\]
\[= a\sqrt{(1- \sin^2 (2t)) }\]

just substitute value of sin2t i juss wanted to see wats happening clearly,thanks @shelby1985

Answer 30

@shubhamsrg

Answer 31

:)

Answer 32

\[a\sqrt{(1-\frac{b^2}{-a^2}})=a\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{a^2+b^2}\]

Answer 33

is this final?

Answer 34

according to me,yes

Answer 35

thanks @asnaseer @shubhamsrg @DLS

Answer 36

yw :)

Answer 37

on this answer wat happened to c^2+d^2 denoinator

Answer 38

In your original problem you got to:\[\begin{align}
x^2-y^2&=\frac{ac+bd}{c^2+d^2}\tag{1}\\
2xy&=\frac{bc-ad}{c^2+d^2}\tag{2}\\
&\text{using the substitutions suggested by shubhamsrg, we get:}\\
x&=\sqrt{\frac{ac+bd}{c^2+d^2}}\cos(t)\tag{3}\\
y&=i\sqrt{\frac{ac+bd}{c^2+d^2}}\sin(t)\tag{4}\\
&\text{this leads to:}\\
\sin(2t)&=\frac{ad-bc}{ac+bd}i\tag{5}\\
\therefore x^2+y^2&=\frac{ac+bd}{c^2+d^2}\sqrt{1-\sin^2(2t))}\\
&=\frac{ac+bd}{c^2+d^2}\sqrt{1+\frac{(ad-bc)^2}{(ac+bd)^2}}\\
&=\frac{ac+bd}{c^2+d^2}\sqrt{\frac{(ac+bd)^2+(ad-bc)^2}{(ac+bd)^2}}\\
&=\frac{\sqrt{(ac+bd)^2+(ad-bc)^2}}{c^2+d^2}\\
&=\frac{\sqrt{a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2}}{c^2+d^2}\\
&=\frac{\sqrt{a^2c^2+b^2d^2+a^2d^2+b^2c^2}}{c^2+d^2}\\
&=\frac{\sqrt{a^2(c^2+d^2)+b^2(c^2+d^2)}}{c^2+d^2}\\
&=\frac{\sqrt{(a^2+b^2)(c^2+d^2)}}{c^2+d^2}\\
&=\sqrt{\frac{a^2+b^2}{c^2+d^2}}\\
\therefore (x^2+y^2)^2&=\frac{a^2+b^2}{c^2+d^2}\\
\end{align}\]

Answer 39

can someone give asneer a medal pls

Answer 40

thank you again

Answer 41

no need for medal?

I am just here to help and learn :)

Answer 42

thank you soo much ,i was wondering wat is keeping you typing for so loooooooong

Answer 43

:) - and thx Joe for the medal :)