How do you solve #sqrt(x^2+2)-5=0#?

2 Answers
Anjali G
Aug 20, 2017

#x=sqrt23, x=-sqrt23#

Explanation:

#sqrt(x^2+2) - 5 = 0#

Add 5 to each side:
#sqrt(x^2+2) = 5#

Square each side:
#x^2+2 =25#

Subtract 2 from each side:
#x^2 = 23#

#color(blue)(x= +- sqrt23)#

smendyka
Aug 20, 2017

See a solution process below:

Explanation:

First, add #color(red)(5)# to each side of the equation to isolate the radical while keeping the equation balanced:

#sqrt(x^2 + 2) - 5 + color(red)(5) = 0 + color(red)(5)#

#sqrt(x^2 + 2) - 0 = 5#

#sqrt(x^2 + 2) = 5#

Next, square each side of the equation to eliminate the radical while keeping the equation balanced:

#(sqrt(x^2 + 2))^2 = 5^2#

#x^2 + 2 = 25#

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

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

#x^2 + 0 = 23#

#x^2 = 23#

Now, take the square root of each side of the equation to solve for #x# while keeping the equation balanced. Remember, the square root of a number produces both a positive and a negative result:

#sqrt(x^2) = +-sqrt(23)#

#x = +-sqrt(23)#

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