Question

what is the extraneous solutions for this problem?
1/x^2+1/x=1/2x^2

Answer 1

I simplified it to x=-1/2 by eliminating the x denominator, multiplying the equation by 2x^2.

Answer 2

is the answer simply no numbers except for x=-1/2, the answer?

Answer 3

If you get common denominators on the LHS and combine, you should get \(\dfrac{1+x}{x^2}=\dfrac{1}{2x^2}\). Cross-multiplying now gives us

\(2x^2(1+x)=1(x^2) \implies 2x^2+x^3=x^2\implies x^2+2x^3=0\implies x^2(1+2x)=0\).

Thus, \(x^2=0\implies x=0\) or \(1+2x = 0\implies x =-\dfrac{1}{2}\).

The extraneous solution in this case would be \(x=0\) since the original equation \(\dfrac{1}{x^2}+\dfrac{1}{x} = \dfrac{1}{2x^2}\) is undefined when \(x=0\).

Does this make sense? :-)

Answer 4

Shoot...left out a two in that large equation; here's the corrected, easier to read version:

\[\begin{aligned}2x^2(1+x)=1(x^2) &\implies 2x^2+\color{red}{2}x^3=x^2\\ &\implies x^2+2x^3=0\\ & \implies x^2(1+2x)=0\end{aligned}\]

Answer 5

But either ways, 0 is the one and only extraneous answer? Thanks for tge help

Answer 6

Yea, 0 is the only extraneous one.

Answer 7

Thanks! so the same would apply for this one too (I think) right?


((x+6)/(4x^2))+((3)/(2x^2))=((x+4)/(2x^2))

Answer 8

Let's see; after getting common denominators on the LHS, the equation simplifies to \(\dfrac{x+12}{4x^2} = \dfrac{x+4}{2x^2} \implies 2x^3+24x^2=4x^3+16x^2\implies 2x^2(x-4)=0\).

(I left a bunch of steps out, but I'm sure you'll be able to fill in what I left out).

So yea, it will be the same exact situation; 0 will again be the only extraneous solution.

Answer 9

Thanks!