Posted by Mellie on Friday, January 7, 2011 at 6:05pm.
The only solution to sqrt (2x+3)  sqrt(x+1) = 1 is x = 3
I think that is true but I'm not sure if I was suppose to separate this equation I just put the 3 in the place of the x and it was true but should I have done it differently?

Algebra  helper, Friday, January 7, 2011 at 6:42pm
The answer is 3 and 1, from Wolframealpha
I arrived at the  1 answer, but not sure if the way I solved it is correct and just luck that I got the correct answer of 1!!
maybe a tutor will step in an answer.

Algebra  Reiny, Friday, January 7, 2011 at 6:44pm
x = 3 does not satisfy the equation as a solution.
LS = (6+3)  √4
= 92
= 7
RS = 1
LS ≠ RS, so x ≠ 3
let's solve it ...
(2x+3)  sqrt(x+1) = 1
2x + 3  1 = √(x+1)
2x+2 = √(x+1)
square both sides
4x^2 + 8x + 4 = x+1
4x^2 + 7x + 3 = 0
(x+1)(4x+3) = 0
x = 1 or x = 3/4
if x = 1
LS = (2+3)  √0
= 10 = 0
= RS
if x = 3/4
LS = 2(3/4) + 3  √(1/4)
= 3/2  1/2 = 1
= RS
so x = 0 or x = 3/4

ARGHH! Algebra  Reiny, Friday, January 7, 2011 at 6:47pm
Forget about my solution above,
Just plain ol' did not see that first square root sign!

2nd try .... Algebra  Reiny, Friday, January 7, 2011 at 6:53pm
√(2x+3) = √(x+1) + 1
square both sides
2x+3 = x+1 + 2√(x+1) + 1
x + 1 = 2√(x+1)
square again ...
x^2 + 2x + 1 = 4(x+1)
x^2  2x  3 = 0
(x+1)(x3) = 0
x = 1 or x = 3
checking both,since we squared.
if x = 1
LS = √(2+3)  √0
= √1√0 = 1 = RS
if x = 3
LS = √9  √4
= 32
= 1 = RS
So x = 1 or x = 3
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