anonymous
  • anonymous
3x^1/2-8x^1/4-3=0 Can you show me how to solve?
Mathematics
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anonymous
  • anonymous
3x^1/2-8x^1/4-3=0 Can you show me how to solve?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
take x^(1/4)=y
anonymous
  • anonymous
solve the quadratic in y.
anonymous
  • anonymous
You are to set up as a substitution problem and solve.

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anonymous
  • anonymous
yes. solve 3y^2-8y-3=0. lastly substitute x^(1/4) for y.
anonymous
  • anonymous
I came up with a substituion of u2-8u-3=0 and came up with 4+-square root 19 but I am not understanding how to substitute the original u=x1/4 and come up with the right answer
anonymous
  • anonymous
its not u^2-8u-3=0 but 3u^2-8u-3=0. if u came with something like u=4+sqrt(19). then x=u^4=(4+sqrt(19))^4. Thats it. :)
anonymous
  • anonymous
Not exactly. You have to substitute back in the x 1/4 for the u. That is where I am stuck and not coming up with the correct answer.
anonymous
  • anonymous
explain where you are stuck.
anonymous
  • anonymous
I solved the quadratics and got u=3,-1/3. then x=3^4=81 or x=(-1/3)^4=1/81.
anonymous
  • anonymous
But x=1/81 does not satisfy the orginal eqn . so our ans is 81
anonymous
  • anonymous
ok. on the original problem I made u=x1/4. At this point the equation should have been 3u^2-8u-3=0. I missed that, you were correct. At that point I plugged it in the quadratic equation. Bottom line I am coming up with 9, 1. So my substituion back should be x1/4=9 and x1/4=1. How do you solve these?
anonymous
  • anonymous
why? x=9^4 or x=1 with ur values.
anonymous
  • anonymous
it is x 1/4 = 9, x 1/4 =1
anonymous
  • anonymous
what is x1/4? Isnt it x^(1/4)?
anonymous
  • anonymous
Im sorry it is x^1/4 =9 and x^1/4 =1
anonymous
  • anonymous
let u=x^1/4 so that u^2 =x^1/2 3u^2 - 8u -3=0 using the quadratic formula u= [-b+ sqrt(b^2-4ac)]/2a
anonymous
  • anonymous
so where are u getting the problem? Raise both sides of the equality (x^1/4=9) to power 4. and u get x=9^4. Is this ur problem?
anonymous
  • anonymous
Yep Mark. That's what I did.
anonymous
  • anonymous
Suabhik, yes. that is my problem.
anonymous
  • anonymous
is it x = 6561 and x=1?
anonymous
  • anonymous
yups bastax.
anonymous
  • anonymous
Well alrighty then. Just talking it through really helped. Thanks saubhik!
anonymous
  • anonymous
hope u r clear with it. :)
anonymous
  • anonymous
i better be. I have a test in an hour!
anonymous
  • anonymous
best of luck!
anonymous
  • anonymous
ok u= [-(-8)+ sqrt((-8)^2-4(3)(-3))]/2*3 u=[8+ sqrt(64+36)]/6 u={8+sqrt(100)/9 u= [8+10]/6 u= 18/6 =3 also the other u is -1/3
anonymous
  • anonymous
Back again on this same question. Book says answer should be 81 and the extraneous is 1/81. How did they get there?
anonymous
  • anonymous
u=x^1/4 =3 x=(3)^4=81
anonymous
  • anonymous
u=(-1/3)^4= 1/81
anonymous
  • anonymous
hope u get the procedure of getting it
anonymous
  • anonymous
and then by using the quadratic formula
anonymous
  • anonymous
I got down to x^1/4=9 and x^1/4 =1. Was this correct? Then how did you come up with the 81, 1/81?
anonymous
  • anonymous
No this is not correct. Please check ur equation.
anonymous
  • anonymous
read the one i wrote on the top one and the second part
anonymous
  • anonymous
let u=x^1/4 so that u^2 =x^1/2 3u^2 - 8u -3=0 using the quadratic formula u= [-b+ sqrt(b^2-4ac)]/2a u= [-(-8)+ sqrt((-8)^2-4(3)(-3))]/2*3 u=[8+ sqrt(64+36)]/6 u={8+sqrt(100)/9 u= [8+10]/6 u= 18/6 =3 also the other u is -1/3
anonymous
  • anonymous
ok. I can see how you got the 3, -1/3. I had made a mistake earlier in the problem.
anonymous
  • anonymous
u=x^1/4 =3 x=(3)^4=81 u=(-1/3)^4= 1/81
anonymous
  • anonymous
ok good luck
anonymous
  • anonymous
Got it!!!! Thank you so much Mark!
anonymous
  • anonymous
ok welcome,,,,, saubnik also got it in his answer...
anonymous
  • anonymous
ok thnx....good luck with our exam
anonymous
  • anonymous
your exam lol
anonymous
  • anonymous
Yep. I saw you both had it! Thanks again!

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