Question

ax^2/x-1= (a+1)^2

Answer 1

what are we solving for?

Answer 2

russian math.im pre russian student.perhaps minus

Answer 3

any body knows?

Answer 4

x = a+1?

Answer 5

try to expand it

Answer 6

i dont know the answer dude

Answer 7

ok ill assume you want to solve for "x" in terms of "a"
\[ax^{2} = (a +1)^{2}(x-1)\]

\[ax^{2}-(a+1)^{2} x+(a+1)^{2} = 0\]
\[x = \frac{(a+1)^{2} \pm \sqrt{(a+1)^{4} -4a(a+1)^{2}}}{2a}\]
\[x = \frac{(a+1)^{2} \pm (a^{2}-1)}{2a}\]

Answer 8

dude,mr rustem said the answer in term x and how you make roots on computer? i dont know

Answer 9

so now you want to solve for "a" in terms of "x" ?
lol

Answer 10

yes.that russian boss.nothing impossible hahaha

Answer 11

i think i made a mistake somewhere anyway....here you go

http://www.wolframalpha.com/input/?i=ax%5E2%2F%28x-1%29%3D+%28a%2B1%29%5E2

Answer 12

dude i have one for u

Answer 13

\[\sqrt{5+\sqrt[3]{x}} + \sqrt{5}\]

Answer 14

wrong question sorry

Answer 15

\[\sqrt{5+\sqrt[3]{x}} + \sqrt{5-\sqrt[3]{x}} = \sqrt[3]{x}\]