1/x^2 + 1/x =6 solve for x

- anonymous

1/x^2 + 1/x =6 solve for x

- jamiebookeater

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- anonymous

make the left side into 1 equation.

- asnaseer

multiply both sides by\(x^2\) and solve the resulting quadratic equation.

- anonymous

\[x^2/ x^2 + x^2/x^2 = 6x^2\]

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## More answers

- anonymous

2x^2 = 6x^2

- asnaseer

the denominator of your second term above is wrong

- asnaseer

you should get:\[x^2/ x^2 + x^2/x = 6x^2\]

- asnaseer

next - simplify each term

- anonymous

but dont they need to have the same denominator to add

- asnaseer

what does \(x^2/x^2\) equal?

- anonymous

1

- asnaseer

correct - now simplify the next term

- anonymous

1+ x= 6x^2

- asnaseer

thats right - now re-arrange this into a standard form, i.e into the form \(ax^2+bx+c=0\) and then solve

- anonymous

so I did quadratic equation and go -1 plus or minus radical 25 all over -12

- asnaseer

doesn't look right - can you please show your steps so that I can help spot where you may have made a mistake?

- anonymous

k, u said put it into the standard form so I started from
1 + x = 6x^2
-6x^2 + x+1 = 0
a= -6, b= 1, c= 1
\[-1\pm \sqrt{1^{2}-4(-6)(1) \over 2(-6)}\]

- anonymous

then got -1 plus or minus radical 25 all over -12

- asnaseer

oh - sorry - I misread

- asnaseer

you are right - and you know what \(\sqrt{25}\) equals?

- asnaseer

the formula you used doesn't look quite right, it should be:\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

- asnaseer

the WHOLE thing should be divided by 2a

- asnaseer

so you should have got:\[\frac{-1\pm \sqrt{1^{2}-4(-6)(1)}}{2(-6)}\]

- anonymous

o ok so um x = 1/-3 and x = 1/2

- asnaseer

perfect!

- anonymous

ya I ment to put all over 2a but I was trying to make it come out nice and pretty in the equation thing but it took lng

- asnaseer

although you should write it as:
x=-1/3 or 1/2

- asnaseer

you don't usually put the minus sign in the denominator

- anonymous

wow these are so frustrating I was literally about to give up and go home, im in a library trying to study and there are all these other students basically hanging out and talking and there are no available tutors

- asnaseer

also note that you should use the word "or" rather than "and" here.
i.e. x = -1/3 or 1/2
instead of x = -1/3 and 1/2
because x canot equal both values at the same time

- anonymous

ok ill do that

- anonymous

thanks a lot you gave me a little hope to keep pushing a little more

- asnaseer

great - and keep persevering - you'll get the hang of these very soon I'm sure :)

- asnaseer

yw :)

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