Question

how to solve the foll set of equations?

Answer 1

and what is that?

Answer 2

10=a^2+b^2

Answer 3

and (9/a^2)-(4/b^2)=1

Answer 4

as a system?

Answer 5

well solve for a and b

Answer 6

show me how

Answer 7

that means as a system...

Answer 8

\[10=a^2+b^2\]\[a^2=10-b^2\]plug that into the other eqn\[{9\over10-b^2}-{4\over b^2}=1\]this should be solvable for b, from which we can get a

Answer 9

but we will get a^4 in denominator ??is there any easier way
?

Answer 10

a quadratic equation in b^2 huh ..

Answer 11

srry b^4

Answer 12

so make it quadrtatic in b^2
that is solve
ax^2+bx+c=0
with x=b^2
then take the sqrt

Answer 13

oh but isnt that lengthy??? any easier way?

Answer 14

use wolframalpha for calculation

Answer 15

Try multiplying the second equation by a^2*b^2. Then multiply the first equation by 4 and add both.

Answer 16

Mani has an interesting idea, I'll check it out :)

Answer 17

ya me too let me check out i knew there was a easy way out

Answer 18

Uh-oh, Sorry folks. You'll still get a quadratic in two variables. :\

Answer 19

:P

Answer 20

\[9b^2-40+4b^2=10b^2-b^4\]\[b^4-b^2-40=0\]let \(x=b^2\)we have\[x^2-x-40=0\]\[(x+4)(x-5)=0\]\[x=\left\{ -4,5 \right\}\]now back to terms of b....

Answer 21

hm... about that -4...
is \(b=2i\) or \(b=\pm2i\) or...
that I'm not sure, but I guess by the fundamental theorem of algebra it must be the latter

Answer 22

so I say\[b=\left\{ -2i,2i,-\sqrt5,\sqrt5 \right\}\]any objections?
I can't say I'm positive everything here is perfect...

Answer 23

oh this is too long?

Answer 24

b=+-2i probably. Because we've to get a total of four solutions from, this biquadratic equation..

Answer 25

oh I didn't factor that equation right
how did you guys let me get away with that?!
I blame you lol

Answer 26

well my text says solving 1 and 2 we get a^2=5 and b^2=5

Answer 27

Which equation?

Answer 28

well the two original eqns lol

Answer 29

a few posts up\[x^2-x-40\neq(x+4)(x-5)\]

Answer 30

lol

Answer 31

no othr way??

Answer 32

arvind I don't think that treating b^2 as x and solving it quadratically is a hard concept, but it looks like the answer is going to be ugly
what does wolf say? let me check

Answer 33

but my text says a^@=5 and b^2=5

Answer 34

well my work agrees that b^2=5, though I'm not sure how...

Answer 35

hmm..its cnfusing

Answer 36

did I make an algebra mistake?
Mani do you get the same thing as me from the sub?

Answer 37

it's not that hard to do the sub...
1)\[10=a^2+b^2\]2)\[\frac9{a^2}-\frac4{b^2}=1\]from 1) we get
3)\[a^2=10-b^2\]which we sub into 2) to get\[\frac9{10-b^2}-\frac4{b^2}=1\]let \(x=b^2\) and we get
4)\[\frac9{10-x}-\frac4{x}=1\]algebra:\[9x-40+4x=x(10-x)\]\[x^2-x-40=0\]make sense so far guys?

Answer 38

ys

Answer 39

so now I guess we can use the quadratic formula or complete the square, eh?
then we can let a^2=y and find a^2 that way
first lets find x I guess,...
i.e. b^2

Answer 40

but i am cnfused with taking double roots

Answer 41

first solving x we get 2 values

Answer 42

let's cross that bridge when we come to it, and for now just find x

Answer 43

then for both the values we need to find a and b

Answer 44

i am ready wt abt u mani??

Answer 45

hmm... there's a prob my mom calling for dinner will be back in 15 min

Answer 46

mani's idea led to the same problem we are having, so only Zarkon knows what we should really be doing
in any event, \(b^2\neq5\) so there must be a typo, or your book is wrong, or...?

Answer 47

dunno i am all confused

Answer 48

actually the qn was like this

Answer 49

you say you know that b^2 is supposed to be 5
and a^2=5 as well ?

Answer 50

yeah pls, what is the \(actual question\)?

Answer 51

find eqn of hyperbola with foci (0+-root10),passing through 2,3

Answer 52

u will see that we reach the 2 eqns

Answer 53

thr i became stuck so i came here

Answer 54

oh I would have to review a bunch of stuff for that...
I really don't know how I can help now, I have class soon

Answer 55

hmm.. this is bad condition i hav no option than leaving that qn