Question

solve x-1=sqrt (x^4-2x^3)/(x^2-4)

Answer 1

Can you take a picture of it please?

Answer 2

The question

Answer 3

Few moments I'll try

Answer 4

ok, thanks

Answer 5

X=2/3

Answer 6

I have to go to work, keep replying please :) I will be back on later tonight

Answer 7

start by factoring the part inside the square root

you get

x^3(x -2)/(x-2)(x+2)

remove the common factor

leaves

x^3/(x + 2)

so square both sides of the equation

(x -1)^2 = x^3/(x +2)

multiply both sides by (x + 2)

and you have

(x+2)(x -1)^2 = x^3

distribute the left had side, collect like terms

and you should have a quadratic to solve

hope it helps

Answer 8

after distributing you get a linear equation... so its easy to solve

Answer 9

Talking about the thing under the square root,
The numerator,x^4-2x^3
Take x^3 common..x^3 (x-2)
And in denominator..use (a^2-b^2)=(a+b)(a-b)
So denominator now is,(x+2)(x-2)
now (x-2) was in numerator too,so both of them cancel out..
we are left with,
x-1=square root of (x^3/x+2)
Square both the sides,
(x-1)^2=(x^3/x+2)
(x+2)(x^2+1-2x)=x^3
x^3+x-2x^2+2x^2+2-4x=x^3
Everything cancels out infact,leaving us with,
3x=2
Therefore,x=2/3

Answer 10

x-1= sqrt (x^4-2x^3)/(x^2-4)
(x-1)^2 = [x^3(x-2)]/[(x-2)(x+2)]
(x^2-2x+1)=x^3/(x+2)
...... * (x+2) = x^3
-3x+2=0
x=2/3

Answer 11

omg, thank you guys SO much :D