Question

Please help! studying for my exam and I don't feel like i am doing it correctly.
#8 a-c
attachment below thanks!

Answer 1

\[\sqrt{3x+1}-x=-1\] if i am reading c correctly
add \(x\) to both sides,
square both sides
solve the resulting quadratic equation
then check your answer

Answer 2

for that one i got x=-1

Answer 3

you get
\[\sqrt{3x+1}=x-1\]
\[3x+1=(x-1)^2\]
\[3x+1=x^2-2x+1\]
set equal to zero
\[x^2-5x=0\]
\[x(x-5)=0\]
\[x=0,x=5\]

Answer 4

then check, you see that 5 works, but 0 does not
clear or no?

Answer 5

yes clear! i did not square both sides thanks.did you happen to look at a?
one of the denominator comes out as an opposite and not sure how to solve that

Answer 6

give me a second

Answer 7

ok since you recognize that one denominator is the negative of one factor of the other denominator, make your life easy and write
\[\frac{1}{x}+\frac{8}{x(x-2)}=\frac{x+2}{x-2}\]

Answer 8

i.e. multiply the second fraction top and bottom by -1, so it becomes and addition
now you have a choice of how to continue,
you can actually add on the left and go from there, or you can multiply both sides of the equal sign by \(x(x-2)\) to clear the fractions
that is probably the easiest method

Answer 9

so x=-3 right?

Answer 10

i got (x-2) (x+3) at the end but it cannot be 2 because 2 was in the denominator right??

Answer 11

yes for sure x cannot be 2
let me check, i didn't actually do the problem

Answer 12

multiplying gives you \(x-2+8=x(x+2)\) or
\[x+6=x^2+2x\]
\[x^2+x-6=0\]
\[(x-2)(x+3)=0\]
\[x=-3\] looks good to me

Answer 13

thank you and can you please guide me through b?