Question

How do you solve `\sqrt { x } + \sqrt { x - 2} < \sqrt { x + 4}`?

Answer 1

`x in [2,(-2+4sqrt(7))/3)`

Explanation:

Given
`color(white)("XXX")sqrt(x)+sqrt(x-2) < sqrt(x+4)`

Note that `x>=2` for `sqrt(x-2)` to be a valid expression.

Since by definition of the square root symbol, all terms are greater than or equal to zero
we can square both sides without changing the validity of the inequality.
`color(white)("XXX")x+2sqrt(x^2-2x)+x-2 < x+4`

`color(white)("XXX")2x-2+2sqrt(x^2-2x) < x+4`

`color(white)("XXX")2sqrt(x^2-2x) < 6-x`

Square both sides again
`color(white)("XXX")4x^2-8x < 36-12x+x^2`

`color(white)("XXX")3x^2+4x-36 < 0`

Using the quadratic formula
`color(white)("XXX")x < (-4+-sqrt(4^2-4 * 3 * (-36)))/(2 * 3)=(-2+-4sqrt(7))/3`

Since we already know that `x >=2` the only valid result is
`color(white)("XXX")x < (-2+4sqrt(7))/3`

...giving the range of values for `x` as in the Answer (above)