Question

How do you solve `\sqrt { x ^ { 2} - 2x + 4} = \sqrt { x ^ { 2} + 5x - 12}`?

Answer 1

`x=2 2/7`

Explanation:

Step 1) Square both sides

`sqrt(x^2-2x+4)=sqrt(x^2+5x-12)`

`x^2-2x+4=x^2+5x-12`

Step 2) Put the variables on the left side and the real numbers on the right

`-2x+4=+5x-12`

`-2x-5x=-12-4`

`-7x=-16`

Step 3) Solve

`x= (-16)/-7`

`x=2 2/7`

Answer 2

`x=16/7=2 2/7`

Explanation:

`sqrt(x^2-2x+4)=sqrt(x^2+5x-12)`

Squaring the two sides

`x^2-2x+4=x^2+5x-12`

i.e. `5x+2x=4+12`

or `7x=16`

i.e. `x=16/7=2 2/7`

Answer 3

`x=16/7`

Explanation:

As there s a square root on each sidez we just square each side to remove the square root:
`(sqrt(x^2-2x+4))^2=(sqrt(x^2+5x-12))^2`

`x^2-2x+4=x^2+5x-12`

As each side has one positive `x^2`, we can just remove them by subtracting `x^2`:

`x^2-2x+4-color(red)(x^2)=x^2+5x-12-color(red)(x^2)`

`-2x+4=5x-12`

Now, we can put the `x`s onto the RHS by adding `2x`:

`-2x+4+color(red)(2x)=5x-12+color(red)(2x)`

`4=7x-12`

Now we just need the integers on the LHS by adding `12`:

`4+color(red)(12)=7x-12+color(red)(12)`

`16=7x`

To get `x` on its own, we just divide both sides by 7:

`16/color(red)(7)=(cancel(7)x)/(cancel(color(red)(7)))`

`x=16/7`