Question 30525
sqrt(2x+5) + sqrt(x+6) = 9----(1)
sqrt(2x+5) = 9 - sqrt(x+6) 
Squaring both the sides
[sqrt(2x+5)]^2 = [9 - sqrt(x+6)]^2
(2x+5) = 9^2+(x+6)-2X9X sqrt(x+6)    
[using [sqrt(p)]^2 = p on the LHS (as well as later in the RHS
and formula: (a-b)^2 = a^2+b^2-2ab where here a=9,b = sqrt(x+6)] 
2x+5 = 81+x+6-18sqrt(x+6)    
(2x-x)+(5-81-6) = -18sqrt(x+6)
(x-82) = -18sqrt(x+6)
Squaring both the sides
(x-82)^2 = [-18sqrt(x+6)]^2
x^2-164x+6724 = 324(x+6)
x^2-164x+6724 = 324x +1944
x^2-164x-324x+6724-1944 =0
x^2-488x+4780 =0
(sum is -488 and productis +4780 and therefore the quantiies are(-478)and (-10))
x^2-10x-478x-4780 = 0
x(x-10)-478(x-10) = 0
xp-478p = 0 where p = (x-10)
p(x-478) = 0
(x-10)(x-478) =0 (putting p back)
(x-10) = 0 gives x = 10
(x-478) = 0 gives x = 478
Answer: x = 10 and x = 478
Verification: x = 10 in (1)
LHS=sqrt(2x+5) + sqrt(x+6) 
=sqrt[(2X10+5)] + sqrt[(10+6)]
= sqrt[(25)] + sqrt[(16)]
=5+4
=9
=RHS
Regarding the other huge number looking at the small number 9 on the RHS tells us that at least one of the roots we should take negative to comply with the answer:
x = 10 in (1)
LHS=sqrt(2x+5) + sqrt(x+6) 
=sqrt[(2X478+5)] + sqrt[(478+6)]
= sqrt[(961)] + sqrt[(484)]
= 31-22
=9
=RHS