Question 31046
My earlier solution is incorrect ,,kindly study this one ,,,

the square root of (3x+10) = 1 + the square root of (2x+5)

on squaring both sdes we get.......

(3x+10) = [1 + the square root of (2x+5)]^2
O
n RHS we apply fromula (a+b)^2 = (a)^2 + (b)^2 + 2(a)(b)
here a = 1 and   b = the square root of (2x+5)..
on applying formala we get ...

(3x+10)=(1)^2 + [the square root of (2x+5)]^2 + 2(1)(the square root of (2x+5)

on solving RHS we get...

3x+10 = 1 + 2x + 5 + 2 (1)(the square root of (2x+5)

3x+10 = 2x + 6 + 2 (1)(the square root of (2x+5)

taking together the like variables we get.....

3x - 2x + 10 - 6  = 2(the square root of (2x+5)

x + 4 = 2(the square root of (2x+5)

again squaring both sides we get

(x + 4)^2 = [2(the square root of (2x+5)]^2

now we apply the earlier fromula on LHS i.e
(a+b)^2 = (a)^2 + (b)^2 + 2(a)(b)
here a = x and b = 4
on applying formula we get...

(x)^2 + (4)^2 + 2(x)(4) = [2(the square root of (2x+5)]^2


x^2 + 16 + 8x = 4(2x+5)

x^2 + 16 + 8x = 8x+ 20

subtract 8x from both sides ,,we get

x^2 + 16  =  20

subtract 16 from both sides ,,we get

x^2 = 4 

taking square root of both sides we get..

either x = 2  0r x = -2


Now if you want to  verify you answer ,,,you can put the values of x in the given equation and if you get LHS=RHS,,,this means that you answer is correct..

hope this will help you 

Please feel free to revert back for any further queries.