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

Question

3717 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-25T17:31:31

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

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

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#

#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.