how do you solve for x? (1/x) + X = 4
multiply everything by x
which x?
both, meant to lower case the second one
:)
adrians got it good then :)
1+x^2=4x, x^2-4x+1=0,
that isnt right though? because you cant factor \[x^2 - 4x + 1\]
its quadratic; so there could be at most 2 answers
don't factor. Solve
well how would you solve it from there?
solutions= -b plus or minus sqrt(b^2-4ac) everything divided by 2a. where a,b and c the coefficients
\[\frac{1}{x} + x = 4\] \[\frac{1}{4x} + \frac{x}{4} = 1\] \[\frac{1}{4x} + \frac{x^2}{4x} = \frac{4x}{4x}\] \[1+x^2 = 4x\] \[x^2-4x+1 = 0\] \[x^2-4x+(4-4)+1 = 0\] \[(x-2)^2-3 = 0\] \[(x-2)^2 = 3\]
can you solve it from there?
how did you go from 1/4x + x/4x = 1/4 to 1/4x + x^2/4x^2 = x/4x?
algebra; its was more work but same conclusion as adrians tho
because if you multiplied both sides by x/x then you would get x/4x^x not 1/4x
you dont have to multiply both sides by "1", thats just a trick used to change form and retain the same value
x/x =1 so it is benign in the process
no what im saying is when you go from the second to the third step in your explination how did you just add an x to the second segment and the right side but not to the first part on the left side?
i was getting like denominators; 4x and 4, i need an x to match; so x/x changes the way it looks without changing the value
oohhh i understand now
the right side was 1/1; so i just modified it to 4x/4x
:)
then how did you take out the 4x on bottom?
too much trouble for a trivial matter. u should multiply and avoid fractions
when you have the same denominators all the way across; the only thing that matters is the tops; you could just as well say that I multiplied all of it by 4x
id have done it adrians way as well; but i like to show that there is more than one way to structure an answer and to use what you feel most comfortable with
alright well thanks i think i understand now :D
dont be afraid to play around with it and see where it takes you :)
haha alright thanks!
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