Question

Solving rational equations.

Answer 1

\[\large \frac{ 1 }{ 2x^2 } = \frac{ 6 }{ x^2 } + \frac{ 1 }{ 4x }\]

Answer 2

@Hero

Answer 3

is the common denominator 4x^3?

Answer 4

4x^2

Answer 5

oh yeah!

Answer 6

if you did that and equate then you break the rationality

Answer 7

it become a nice linear equation

Answer 8

All rational equations share the same methodology in terms of solutions, this methodology being first trying to represent the variable on the numerator to get a potential solution.
So, having:
\[\frac{ 1 }{ 2x^2 }=\frac{ 6 }{ x^2 }+\frac{ 1 }{ 4x }\]
I will suppose you can operate fractions, so, we operate the fractions on the right handside in order to make everything more simple:
\[\frac{ 1 }{ 2x^2 }=\frac{ 6(4x)+x^2 }{ 4x^3 }\]
This is what is called a "proportion" since we have two fractions separated by a "=" sign, we can therefore cross multiply: \[\frac{ 1 }{ 2x^2 }=\frac{ 24x+x^2 }{ 4x^3 } \iff 4x^3=(2x^2)(24x+x^2) \]
And then applying distributive property in order to get rid of the parenthesis:
\[4x^3=(2x^2)(24x+x^2) \iff 4x^3=48x^3+2x^4\]
And this is good enough to make it all equal zero, and solve for more easily:
\[4x^3=48x^3+2x^4 \iff (48-4)x^3+2x^4=0 \]
Now it became easier to solve, we transformed the rational equation into a very simple fourth degree equation which you should be able to solve at this point, I'll leave the rest to you.

Answer 9

i got x = -22

Answer 10

well there different ways of approaching this!
however the way you looking for is using 4x^2
\[\frac{2}{4x^2}=\frac{24}{4x^2}+\frac{x}{4x^2}\]
\[2=24+x \Longrightarrow x=-22\]

Answer 11

@Owlcoffee, has one flow
x cannot be 0
if you solve that final equation you need to pay attention the fact that the domain excludes 0

Answer 12

That's called "evaluating the solution", since polynomial equations do not present any problems in the domain it is not performed.
In the case of radical or rational equations it MUST be done, since not all the x-values that appear as a "solution" are actually solutions.