# What is X if X-6X^(1/2)=-8

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We need to determine the value of `x ` in `x-6sqrt(x)=-8`

First, we isolate the term containing square root. So we have

`6sqrt(x) =x+8`

Then if we square both sides, we get

`36x=(x+8)^2`

By using FOIL method and further simplifying

`36x = x^2+16x+64`

`x^2-20x+64=0`

By factoring,

`(x-16)(x-4)=0`

By zero product property, the solution in the given equaiton is eitherĀ `x=16` or `x=4` .

The equation `x-6x^(1/2)=-8` has to be solved for x.

Let `x^(1/2)=t` , then `x=t^2`

The given equation reduces to:

`t^2-6t=-8`

`rArr t^2-6t+8=0`

`rArr t^2-4t-2t+8=0`

`rArr t(t-4)-2(t-4)=0`

`rArr (t-4)(t-2)=0`

To solve for `t` , put each of the terms equal to zero.

`(t-4)=0`

`rArr t=4`

Also, `(t-2)=0`

`rArr t=2`

When `t=4, x=t^2`

`=4^2`

`=16`

When `t=2, x=t^2`

`=2^2`

`=4`

Therefore, the values of x for which the equation `x-6x^(1/2)=-8` holds good are 2 and 4.