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

2 Answers

Answer:

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

Answer:

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#