Question

Solve for x -

2(x^2 + 1/x) -3 (x-1/x) -4 =0

I remember that when I earlier did this i took something as another variable and then got a quadratic equation from which I solved for the possible values of the taken variable !!! Help me please guys !!!!

Answer 1

Is this your question
\[2(x^2+ \frac{1}{x})-3 (x-\frac{1}{x})-4=0\]

Answer 2

yes absolutely right

Answer 3

I think the first term should be
\[2 (x^2+\frac{1}{x^2})\]
you sure about this?

Answer 4

oops sorry !! it is x^2 +1/X^2 sorry !! you r correct

Answer 5

Great :D now Let's solve this

Answer 6

Anybody Help please !!!!

Answer 7

we have
\[2(x^2+\frac{1}{x^2}) -3(x-\frac{1}{x})-4=0\]
we know that
\[(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2\]
so
\[(x-\frac{1}{x})^2+2=x^2+\frac{1}{x^2}\]
so we have in our equation
\[2((x-\frac{1}{x})^2+2) -3(x-\frac{1}{x})-4=0\]
Let's substitute
\[x-\frac{1}{x}= t\]
so we get
\[2(t^2+2)-3t-4=0\]
we get now
\[2t^2+4-3t-4=0\]
we get
\[2t^-3t=0\]
now
we get
\[t(2t-3)=0\]
so either t=0 or t= 3/2
\[x-\frac{1}{x}= 0 or 3/2\]
Can you solve now

Answer 8

Ok great thanks a lot now I remember !!!!!

Answer 9

we have
\[x-\frac{1}{x}=0\]
so
\[x^2-1=0=> x=\pm 1\]
or
\[x-1/x=\frac{3}{2}\]
so we get
\[2x^2-2=3x\]
we get
\[2x^2-3x-2=0\]
so
\[2x^2-4x+x-2=0\]
\[2x(x-2)+1(x-2)=0\]
so
\[x=2 or x=-\frac{1}{2}\]
so x=1 or -1 or -1/2 or 2

Answer 10

Ok got it already just needed the first 3 steps !!!! still thanks a lot !!!