Question

x=(3/x)+(1/2) ??

Answer 1

Do you want to solve for x?

Answer 2

yes

Answer 3

i know what it is

Answer 4

\[x=\frac{ 3 }{ x }+\frac{ 1 }{ 2 } \rightarrow x=\frac{ 6+x }{ 2x } \rightarrow 2x ^{2}-6-x=0\]

Answer 5

So we have:
\[x=(\frac{ 3 }{ x })+(\frac{ 1 }{ 2 })\]
What we have to do first is something called "common denominator" wich can allow us to sum those fractions:
So applying it, we get:
\[x=\frac{ 6+x }{ 2x }\]
Now we can multiply both sides by "2x" to get rid of that "2x" in the denominator:
\[(2x)x=6+x\]
I'll operate that multiplication:
\[2x ^{2}=6+x\]
I'll substract 6 and x to both sides:
\[2x ^{2}-x-6=0\]
Now this has the form:
\[ax ^{2}+bx+c=0\]
meaning that we can solve this using the general formula:
\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]
I encourage you to apply it.

Answer 6

factor and find the zeros. then check in the original

Answer 7

easier to factor

Answer 8

So x could be either 2 or -3/2? @Owlcoffee

Answer 9

yep

Answer 10

The general formula for quadratic equations will always throw one or two solutions.
Just in some cases we can say that there are no solutions.
But those values are correct.