How do you solve #sqrt(2x+10)-2sqrtx=0#?

2 Answers
Jul 17, 2016

Answer:

#x = 5#

Explanation:

Add #2sqrt(x)# to both sides:

#sqrt(2x+10) = 2sqrt(x)#

Square both sides:

#2x+10 = 4x#

Subtract #2x# from both sides:

#10 = 2x#

#implies x = 5#

Jul 17, 2016

Answer:

#color(blue)(5 = x)#

Explanation:

First get #-2sqrtx# to the other side.

#sqrt(2x+10) cancel(- 2sqrtx + 2sqrtx) = 0 + 2sqrtx# - use additive inverse

#sqrt(2x+10) = 2sqrtx#

Now square both sides. Remember that squaring a square root is equal to that number under the square root.

#(sqrt(2x+10))^2 = (2sqrtx)^2# - square each side, because what you do to one side, you must do to the other

#2x+10 = 2^2 * (sqrt(x))^2# - Follow this concept: [#(ab)^x = a^x * b^x#]

#2x+10 = 4x#

Now we can isolate the variable and identify #x#. Get #2x# to the other side and combine it with #4x#, then divide out the coefficient to find #x#.

#cancel(2x-2x) + 10 = 4x-2x# - use additive inverse

#10 = 2x# - combine like terms

#10/2 = (cancel(2)x)/(cancel(2))# - divide by two

Final Answer:

#color(blue)(5 = x)#