Given: #sqrt(2x+4)+sqrt(3x-5)=4" [1]"#
If we mulitply both sides by the conjugate, #sqrt(2x+4)-sqrt(3x-5)#, it will eliminate the radicals from the left side:
#2x+4-(3x-5)=4(sqrt(2x+4)-sqrt(3x-5))" [2]"#
Combine like terms and distribute the 4:
#9-x=4sqrt(2x+4)-4sqrt(3x-5)" [3]"#
Flip equation [3]:
#4sqrt(2x+4)-4sqrt(3x-5)= 9-x" [4]"#
Multiply both sides of equation [1] by 4:
#4sqrt(2x+4)+4sqrt(3x-5)=16" [5]"#
We can make the term #4sqrt(3x-5)# disappear by adding equation [4] to equation [5]:
#8sqrt(2x+4)=25-x" [6]"#
Square both sides:
#64(2x+4) = x^2 - 50x + 625#
Distribute the 64:
#128x+256 = x^2 - 50x + 625#
Combine like terms:
#x^2 - 178x + 369=0#
Using the quadratic formula:
#x = 89-8sqrt(118)# and #x=89+8sqrt(118)#
Check both:
#sqrt(2(89-8sqrt(118))+4)+sqrt(3(89-8sqrt(118))-5)=4#
#4 = 4#
#sqrt(2(89+8sqrt(118))+4)+sqrt(3(89+8sqrt(118))-5)=4#
#41.71!=4#
We must discard the second root; the only solution is:
#x = 89-8sqrt(118)#
