How do you solve #sqrt(2x+10)-2sqrtx=0#?
2 Answers
Explanation:
Add
Square both sides:
Subtract
Explanation:
First get
#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
#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)#