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.
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?
4 answers
x = 3 does not satisfy the equation as a solution.
LS = (6+3) - √4
= 9-2
= 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
= 1-0 = 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
LS = (6+3) - √4
= 9-2
= 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
= 1-0 = 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
Forget about my solution above,
Just plain ol' did not see that first square root sign!
Just plain ol' did not see that first square root sign!
√(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)(x-3) = 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
= 3-2
= 1 = RS
So x = -1 or x = 3
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)(x-3) = 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
= 3-2
= 1 = RS
So x = -1 or x = 3