Asked by Mellie
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?
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?
Answers
Answered by
helper
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.
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.
Answered by
Reiny
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
Answered by
Reiny
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!
Answered by
Reiny
√(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
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