Question

How do you solve `sqrt( x -10) = 5x`?

Answer 1

There are no Real solutions,
if Complex values are allowed `x=(1+-isqrt(999))/50`

Explanation:

Given
`color(white)("XXX")sqrt(x-10)=5x`

Working in the domain of real numbers we must ensure that

`x>=10` and `x>=0` hence finally `x>=10`.

Squaring both sides:
`color(white)("XXX")x-10=25x^2`

Rewriting in standard quadratic form:
`color(white)("XXX")25x^2-x+10=0`

Applying the quadratic formula
`color(white)("XXX")x=(1+-sqrt(-999))/50`

Because Real numbers do not permit square roots of negative values there are no Real solutions.

Answer 2

We have complex roots only which are `x=1/50+-isqrt(999)/50`

Explanation:

Squaring the two sides in `sqrt(x-10)=5x`, we get

`x-10=25x^2`

or `25x^2-x+10=0`

according to quadratic formula, the solution of the

equation `ax^2+bx+c=0` is given by

`x=(-b+-sqrt(b^2-4ac))/(2a)`.

Hence, solution is `x=(-(-1)+-sqrt((-1)^2-4*25*10))/(2*25)`

or `x=(1+-sqrt(1-1000))/(50)=(1+-sqrt(-999))/50`

As discriminant is negative, we only have complex roots,

which are `x=(1+-isqrt(999))/50`