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

2 Answers

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.

May 31, 2016

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