Question

How to solve for `x` if `x^(2/3)-5x^(1/3)+6=0` ?

Answer 1

`x=8` or `x=27`

Explanation:

`x^(2/3)-5x^(1/3)+6=0`

Let `y=x^(1/3)`, and then the equation becomes a quadratic in `y`

`y^2-5y+6=0`

`(y-2)(y-3)=0`

`y-2=0 rArr x^(1/3)=2 rArr x=8`

`y-3=0rArrx^(1/3)=3rArrx=27`

Answer 2

Solution: `x=8 ,x=27`

Explanation:

`x^(2/3)-5x^(1/3)+6=0`. Let `x^(1/3)=a :.x^(2/3)=a^2`

`a^2 -5a+6=0 or a^2-2a-3a+6=0` or

`a(a-2)-3(a-2)=0 or (a-2)(a-3)=0`

Either `a-2=0 :. a=2` or `a-3=0 :. a=3`

When `a=3; x^(1/3)=3`. Cubing both sides we get `x=27`

When `a=2; x^(1/3)=2`. Cubing both sides we get `x=8`

Solution: `x=8 ,x=27` [Ans]

Answer 3

`x=27`, or `x=8`

Explanation:

First get the equation into a form that is easier to solve, ie. `ax^2+bx+c`. So:

`x^(2/3)+5x^(1/3)+6=0`

`=>x^(2\times1/3)+5x^(1/3)+6=0`

`=>x^((1/3)^2)+5x^(1/3)+6=0` `color(white)(blank)`exponent laws `(x^a\timesb=x^((a)^b))`

You can see that it is pretty close to the form `ax^2+bx+c`. It will get to this form if it were `x` instead of `x^(1/3)`. So let `x^(1/3)=u`

`=>u^2+5u+6=0` `color(white)(blank)`letting `x^(1/3)=u`

`=>(u-3)(u-2)=0`

`=>u-3=0` `color(white)(blankspae)`or `color(white)(blankspace)` `u-2=0`

`=>u=3` `color(white)(blankspaceeee)`or `color(white)(blankspace)` `u=2`

Now we can substitute `u` for `x^(1/3)`

`=>x^(1/3)=3` `color(white)(blankspaceee)`or `color(white)(blankspace)` `x^(1/3)=2`

`=>x=3^3` `color(white)(blankspaceee)`or `color(white)(blankspace)` `x=2^3`

`=>x=27` `color(white)(blankspaceee)`or `color(white)(blankspace)` `x=8`