How do you solve #45-sqrt(10-2x^2)=25#?

1 Answer
Jun 29, 2017

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(45)# from each side of the equation to isolate the radical term while keeping the equation balanced:

#-color(red)(45) + 45 - sqrt(10 - 2x^2) = -color(red)(45) + 25#

#0 - sqrt(10 - 2x^2) = -20#

#-sqrt(10 - 2x^2) = -20#

Next, square both sides of the equation to eliminate the radical while keeping the equation balanced:

#(-sqrt(10 - 2x^2))^2 = (-20)^2#

#10 - 2x^2 = 400#

Then, subtract #color(red)(10)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-color(red)(10) + 10 - 2x^2 = -color(red)(10) + 400#

#0 - 2x^2 = 390#

#-2x^2 = 390#

Then, divide each side of the equation by #color(red)(-2)# to isolate #x^2# while keeping the equation balanced:

#(-2x^2)color(red)(-2) = 390/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))x^2)cancel(color(red)(-2)) = -195#

#x^2 = -195#

Because any number squared always produces a positive result, there is no solution for #x# which will result in a negative 195.

Or, the solution is the null or empty set: #{O/}#