How do you solve #-sqrtx = -25x#?

1 Answer
Aug 5, 2015

Answer:

#x = 0# and #x= 1/625#

Explanation:

First, rewrite your equation by cancelling the minus sign

#sqrt(x) = 25x#

Right from the start, it's obvious that any solution you find must be positive, #x>=0#, since the square root of a negative number does not have a real number as a solution.

So, square both sides of the equation to get

#(sqrt(x))^2 = (25x)^2#

#x = 625x^2#

Move everything to one side of the equation and factor the resulting expression

#625x^2 - x = 0#

#x(625x - 1) = 0#

This equation is equal to zero when wither #x=0# or when #(625x-1)=0#, which is equivalent to having

#625x-1 = 0 => x= 1/625#

Both solutions are valid because they satisfy the condition #x>=0#.