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anonymous
 4 years ago
1/x^2 + 1/x =6 solve for x
anonymous
 4 years ago
1/x^2 + 1/x =6 solve for x

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0make the left side into 1 equation.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0multiply both sides by\(x^2\) and solve the resulting quadratic equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x^2/ x^2 + x^2/x^2 = 6x^2\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0the denominator of your second term above is wrong

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0you should get:\[x^2/ x^2 + x^2/x = 6x^2\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0next  simplify each term

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but dont they need to have the same denominator to add

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0what does \(x^2/x^2\) equal?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0correct  now simplify the next term

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0thats right  now rearrange this into a standard form, i.e into the form \(ax^2+bx+c=0\) and then solve

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so I did quadratic equation and go 1 plus or minus radical 25 all over 12

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0doesn't look right  can you please show your steps so that I can help spot where you may have made a mistake?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k, 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
 4 years ago
Best ResponseYou've already chosen the best response.0then got 1 plus or minus radical 25 all over 12

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0oh  sorry  I misread

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0you are right  and you know what \(\sqrt{25}\) equals?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0the formula you used doesn't look quite right, it should be:\[x=\frac{b\pm\sqrt{b^24ac}}{2a}\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0the WHOLE thing should be divided by 2a

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0so you should have got:\[\frac{1\pm \sqrt{1^{2}4(6)(1)}}{2(6)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0o ok so um x = 1/3 and x = 1/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya 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
 4 years ago
Best ResponseYou've already chosen the best response.0although you should write it as: x=1/3 or 1/2

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0you don't usually put the minus sign in the denominator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wow 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
 4 years ago
Best ResponseYou've already chosen the best response.0also 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
 4 years ago
Best ResponseYou've already chosen the best response.0thanks a lot you gave me a little hope to keep pushing a little more

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0great  and keep persevering  you'll get the hang of these very soon I'm sure :)
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