Question

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

Answer 1

take x^(1/4)=y

Answer 2

solve the quadratic in y.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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. :)

Answer 7

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.

Answer 8

explain where you are stuck.

Answer 9

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

Answer 10

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

Answer 11

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?

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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?

Answer 18

Yep Mark. That's what I did.

Answer 19

Suabhik, yes. that is my problem.

Answer 20

is it x = 6561 and x=1?

Answer 21

yups bastax.

Answer 22

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

Answer 23

hope u r clear with it. :)

Answer 24

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

Answer 25

best of luck!

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

hope u get the procedure of getting it

Answer 31

and then by using the quadratic formula

Answer 32

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?

Answer 33

No this is not correct. Please check ur equation.

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

ok good luck

Answer 39

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

Answer 40

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

Answer 41

ok thnx....good luck with our exam

Answer 42

your exam lol

Answer 43

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