Method 4. In other words, how I'd do it if I didn't spot the diff of squares. sqrt(x + a) + sqrt(x) = a sqrt(x + a) - sqrt(x) = b Subtract the two: 2 sqrt(x) = a - b Divide through by 2 sqrt(x) = (a - b)/2 Square it x = (a^2 - 2ab + b^2)/4 Add the initial two: 2 sqrt(x + a) = a + b Divide through by 2 sqrt(x + a) = (a + b)/2 Square it x + a = (a^2 + 2ab + b^2)/4 Substituting in the x from the subtraction step: (a^2 - 2ab + b^2)/4 + a = (a^2 + 2ab + b^2)/4 Multiplying through by 4 a^2 - 2ab + b^2 + 4a = a^2 + 2ab + b^2 Subtract a^2 + b^2 from both sides -2ab + 4a = 2ab Add 2ab to both sides 4a = 4ab Either a = 0 or b = 1. Well, we wanted b. So, b = 1, unless a = 0. Let's check that cases: sqrt(x + 0) + sqrt(0) = 0 sqrt(x) = 0 x = 0 sqrt(0 + 0) - sqrt(0) = 0 - 0 = 0. So, if a = 0, then b = 0. Otherwise, b = 1.
simple: for a =/= 0, (sqrt(x+a) + sqrt(x))(sqrt(x+a) - sqrt(x)) = x + a - x a(sqrt(x+a) - sqrt(x)) = a (sqrt(x+a) - sqrt(x)) = 1 for a = 0, x = 0 so (sqrt(x+a) - sqrt(x)) = 0
I just multiplied sqrt(x + a) + sqrt(x) = a by sqrt(x + a) - sqrt(x) on both sides to get: a = a(sqrt(x + a) - sqrt(x) ) Clearly if a != 0, sqrt(x + a) - sqrt(x) = 1
From (A-B)(A+B)=A^2-B^2, first expression times the second = a. Hence the second is 1. Another presentation by cybermath that is long-winded and trivial.
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Inversed: 1/(sqrt(x+a)+sqrt(x)) = 1/a
multiply sqrt(x+a)-sqrt(x) to left, the result is 1
Method 4. In other words, how I'd do it if I didn't spot the diff of squares.
sqrt(x + a) + sqrt(x) = a
sqrt(x + a) - sqrt(x) = b
Subtract the two:
2 sqrt(x) = a - b
Divide through by 2
sqrt(x) = (a - b)/2
Square it
x = (a^2 - 2ab + b^2)/4
Add the initial two:
2 sqrt(x + a) = a + b
Divide through by 2
sqrt(x + a) = (a + b)/2
Square it
x + a = (a^2 + 2ab + b^2)/4
Substituting in the x from the subtraction step:
(a^2 - 2ab + b^2)/4 + a = (a^2 + 2ab + b^2)/4
Multiplying through by 4
a^2 - 2ab + b^2 + 4a = a^2 + 2ab + b^2
Subtract a^2 + b^2 from both sides
-2ab + 4a = 2ab
Add 2ab to both sides
4a = 4ab
Either a = 0 or b = 1.
Well, we wanted b. So, b = 1, unless a = 0. Let's check that cases:
sqrt(x + 0) + sqrt(0) = 0
sqrt(x) = 0
x = 0
sqrt(0 + 0) - sqrt(0) = 0 - 0 = 0.
So, if a = 0, then b = 0. Otherwise, b = 1.
this problen has sense only for a=0, and a>1, a=1.
Method 3 please. LOL
simple:
for a =/= 0,
(sqrt(x+a) + sqrt(x))(sqrt(x+a) - sqrt(x)) = x + a - x
a(sqrt(x+a) - sqrt(x)) = a
(sqrt(x+a) - sqrt(x)) = 1
for a = 0, x = 0 so (sqrt(x+a) - sqrt(x)) = 0
The solution is valid for a>1.
Otherwise, sqrt(x) = (a-1)/2 is negative.
I just multiplied sqrt(x + a) + sqrt(x) = a by sqrt(x + a) - sqrt(x) on both sides to get:
a = a(sqrt(x + a) - sqrt(x) )
Clearly if a != 0, sqrt(x + a) - sqrt(x) = 1
Do we need to consider the absolute value for √ (·)² appearing in method 1 and 2 when there’s no specification on parameter a?
Absolutely.
Answer:
√(x+a) - √x = 1 OR √(x+a) - √x = 0
Calculation:
√(x+a) + √x = a
√(x+a) - √x = ?
(√(x+a) + √x) * (√(x+a) - √x) =
= (√(x+a))² - (√x)²
= (x+a) - x
= a
==>
(√(x+a) + √x) * (√(x+a) - √x) = a
and
(√(x+a) + √x) = a
==>
a * (√(x+a) - √x) = a
a * (√(x+a) - √x) - a = 0
a * [ (√(x+a) - √x) - 1 ] = 0
[ (√(x+a) - √x) - 1 ] = 0 OR a = 0
(√(x+a) - √x) = 1 OR a = 0
(√(x+a) - √x) = 1 OR (√(x+a) - √x) = (√(x+0) - √x) = (√x - √x) = 0
(a-√x)² = x+a
a²-2a√x-a=0
a [a-(2√x+1)] = 0
a =0, a = 2√x+1
a=0 implies x=0
a = 2√x+1
√(x+a) -√x =
√(x+2√x+1) -√x =
√[(√x+1 )² ]-√x =
√x+1 - √x =
1
answer
√(x+a) -√x = 1
1
From (A-B)(A+B)=A^2-B^2, first expression times the second = a. Hence the second is 1. Another presentation by cybermath that is long-winded and trivial.