Question

How do you solve `\root[ 4] { 3- 50x ^ { 2} } = 5x`?

Answer 1

From the outset, it might be helpful to note that solutions will only be valid when `3-50x^2>0`, since a fourth root must be positive. That inequality can be factored as `(sqrt3-5sqrt2x)(sqrt3+5sqrt2x)>0` which is true on `-5sqrt(2/3)ltxlt5sqrt(2/3)`.

Also note that if `x<0`, when the right hand side of the equation is negative. However, the left-hand side of the equation can only be positive, so we also have the condition `x>=0`.

Thus our solution(s) must fall within `0lt=x<5sqrt(2/3)`.

Going through this process isn't necessary, but it eliminates having to plug in our answers at the end to see if they've been introduced extraneously or not.

To undo a `4`th root, raise both sides to the `4`th power.

`root(4)(3-50x^2)=5x`

`(root4(3-50x^2))^4=(5x)^4`

`3-50x^2=625x^4`

`625x^4+50x^2-3=0`

Let `t=x^2`:

`625t^2+50t-3=0`

Applying the quadratic formula:

`t=(-50+-sqrt(50^2-4(625)(-3)))/(2(625))`

`t=(-50+-100)/(2(625))`

`t=-3/25,1/25`

`x^2=-3/25,1/25`

The negative solution is discarded, leaving `x^2=1/25`, so:

`x=+-1/5`

As per the discussion before, the negative answer is invalid.
Note that `5sqrt(2/3)approx4.08`, so `x=1/5` is well within our previously defined domain. It is the only solution:

`x=1/5`