Question

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

Answer 1

`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`

Answer 2

`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)`