3x^1/2-8x^1/4-3=0 Can you show me how to solve?

- anonymous

3x^1/2-8x^1/4-3=0 Can you show me how to solve?

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- anonymous

take x^(1/4)=y

- anonymous

solve the quadratic in y.

- anonymous

You are to set up as a substitution problem and solve.

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## More answers

- anonymous

yes. solve 3y^2-8y-3=0. lastly substitute x^(1/4) for y.

- 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

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

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

explain where you are stuck.

- anonymous

I solved the quadratics and got u=3,-1/3. then x=3^4=81 or x=(-1/3)^4=1/81.

- anonymous

But x=1/81 does not satisfy the orginal eqn . so our ans is 81

- 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

why? x=9^4 or x=1 with ur values.

- anonymous

it is x 1/4 = 9, x 1/4 =1

- anonymous

what is x1/4? Isnt it x^(1/4)?

- anonymous

Im sorry it is x^1/4 =9 and x^1/4 =1

- 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

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

Yep Mark. That's what I did.

- anonymous

Suabhik, yes. that is my problem.

- anonymous

is it x = 6561 and x=1?

- anonymous

yups bastax.

- anonymous

Well alrighty then. Just talking it through really helped. Thanks saubhik!

- anonymous

hope u r clear with it. :)

- anonymous

i better be. I have a test in an hour!

- anonymous

best of luck!

- 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

Back again on this same question. Book says answer should be 81 and the extraneous is 1/81. How did they get there?

- anonymous

u=x^1/4 =3
x=(3)^4=81

- anonymous

u=(-1/3)^4= 1/81

- anonymous

hope u get the procedure of getting it

- anonymous

and then by using the quadratic formula

- 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

No this is not correct. Please check ur equation.

- anonymous

read the one i wrote on the top one and the second part

- 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

ok. I can see how you got the 3, -1/3. I had made a mistake earlier in the problem.

- anonymous

u=x^1/4 =3
x=(3)^4=81
u=(-1/3)^4= 1/81

- anonymous

ok good luck

- anonymous

Got it!!!! Thank you so much Mark!

- anonymous

ok welcome,,,,, saubnik also got it in his answer...

- anonymous

ok thnx....good luck with our exam

- anonymous

your exam lol

- anonymous

Yep. I saw you both had it! Thanks again!

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