Question

x^3-9x^3/2+8=0

Answer 1

Do you mean:
\[x ^{3} - \frac{ 9x ^{3} }{ 2 } + 8 = 0\]

Answer 2

\[x ^{3}-9x ^{(3/2)}+8=0\]

Answer 3

@shamim yeah, that's what I mean

Answer 4

\[or, x ^{3}-8x ^{(3/2)}-x ^{(3/2)}+8=0\]

Answer 5

\[x ^{(3/2)}(x ^{(3/2)}-8)-(x ^{(3/2)}-8)=0\]

Answer 6

\[or, (x ^{(3/2)}-1)(x ^{3/2}-8)=0\]

Answer 7

\[I got this far (u-8)(u-1)=0 then u=8 u=1 \]

Answer 8

ok

Answer 9

It says the answer is x=1and x=4 but I was wondering how the 4 came about.

Answer 10

\[\let x ^{3/2}=u\]

Answer 11

ok i m now trying to clearifying ur problem

Answer 12

This is what I did: u^2-9u+8=0 then (u-8)(u-1)=0 factored. so u=8 and u=1 but then how did the text get x=4?

Answer 13

ok

Answer 14

let \[x ^{3/2}=u\]

Answer 15

then our equation will b
\[u ^{2}-9u+8=0\]

Answer 16

\[or, u ^{2}-8u-u+8=0\]

Answer 17

\[or, u(u-8)-(u-8)=0\]

Answer 18

\[or, (u-1)(u-8)=0\]

Answer 19

\[u-1=0 \]

Answer 20

\[or, u=1\]

Answer 21

\[or, x ^{3/2}=1\]