Question

How do you solve `sqrt(x+5) + sqrt(x+15) = sqrt(9x+40)`?

Answer 1

Square, rearrange and square again to get a quadratic, one of whose roots is a valid solution:

`x = (-20+2sqrt(55))/9 ~= -0.574`

Explanation:

Note that both sides are non-negative, being the sums of non-negative square roots. So squaring both sides will not introduce spurious solutions.

So square both sides to get:

`9x+40 = (x+5) + 2sqrt(x+5)sqrt(x+15)+(x+15)`

`=2x+20+2sqrt(x^2+20x+75)`

Subtract `2x+20` from both ends to get:

`7x+20 = 2sqrt(x^2+20x+75)`

Note that a valid solution will require `7x + 20 >= 0`.

Square both sides (which may introduce extraneous solutions with `7x+20 < 0`) to get:

`49x^2+280x+400 = 4x^2+80x+300`

Subtract the right hand side from the left to get:

`45x^2+200x+100 = 0`

Divide through by `5` to get:

`9x^2+40x+20 = 0`

Solve using the quadratic formula:

`x = (-40+-sqrt(40^2-(4xx9xx20)))/(2*9)`

`=(-40+-sqrt(1600-720))/18`

`=(-40+-sqrt(880))/18`

`=(-40+-sqrt(16*55))/18`

`=(-40+-4sqrt(55))/18`

`=(-20+-2sqrt(55))/9`

`(-20-2sqrt(55))/9 ~= -3.870`

`7(-3.870)+20 < 0` so this solution is spurious.

`(-20+2sqrt(55))/9 ~= -0.574`

`7(-0.574)+20 > 0` so this solution is valid.