Evaluating a Radical Expression in Three Ways

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[armacham]

You can also get rid of radicals by adding and subtracting the original expressions, to get:
2sqrt(x+a) = a + u
2sqrt(x) = a – u
Then you can square both sides
4(x+a) = a^2 + 2au + u^2
4(x) = a^2 – 2au + u^2
Subtract the second equation from the first to get:
4a = 4au
0 = a(u – 1)
Either a = 0 or u = 1
And if you test a=0, you get a=0, x=0, u=0

[gorjaharchangel]

I immediately thought of the last method. It was quite obviously a difference of squares

[GirishManjunathMusic]

Given:
√(x + a) + √x = a
To find:
y = √(x + a) – √x

Mutlipying:
(√(x + a) + √x)·(√(x + a) – x) = a·y

Using (√r + √q)·(√r – √q) = r – q:
x + a – x = ay
a = ay

Assuming a ≠ 0:
y = 1.

If a = 0, then x also has to be 0. In this case, y = 0 + 0 – 0 = 0.

[andylee]

Lesson learned: I always need to keep an eye out for conjugates!

The first two methods weren't too painful, but of course it's nice to spot the most elegant way.

[jmart]

I could see 3rd method very quickly.
Very funny. Thanks!

[chrissekely]

Just multiply both equations with each other. After simplifying the results, you'll have x + a - x = a * ? (using ? as a variable). This results in a = a * ?, so ? = 1.

[gdtargetvn]

The fourth method is that
a can be rewritten as:
a = (x + a) - x = [√(x + a) + √x] • [√(x + a) - √x] = a • [√(x + a) - √x]

From here, it's clear that the answer is 1
(Also if a = 0 then the answer is 0)

[허공]

let sqrt(x+a)=b, sqrt(x)=c then x+q=b^2, x=c^2 and b^2-c^2=a
given eq. sqrt(x+a)+sqrt(x)=a, b+c=a
b+c=a, b^2-c^2=(b+c)(b-c)=a. if a is not equal to 0 then b-c=1
which means sqrt(x+a)-sqrt(x)=1.

[geoffreyparfitt]

What I find curious is that whenever x is any square number, a can be solved as an integer. Of course any of those integer pairs (x, a) can be substituted to achieve the unit solution.

[ManjulaMathew-wbzn]

Add the equations and square 4x+4a=(a+u)^2
Subtract equations and square. 4x=(a-u)^2
Subtract the two equations 4a=4au then u=1

[qwang]

If a = 0, then sqrt(x + a) - sqrt(x) = 0
Assume a <> 0. Write sqrt(x + a) - sqrt(x) = a - 2*sqrt(x).
Also write sqrt(x+a) = a - sqrt(x). Taking sq, x + a = a^2 + x - 2a*sqrt(x).
So -2*sqrt(x) = 1 - a. sqrt(x + a) - sqrt(x) = a - 2*sqrt(x) = 1.
..

[BackroomsFan-wdwf]

I managed to find a very elegant solution to this which only requires 3 steps.

So we have the equation sqrt(x+a) + sqrt(x) = a and we want to solve for sqrt(x+a) - sqrt(x).

Now assuming that the second equation returns a constant or any other equation with respect to x, let’s call it y. This way we have:

sqrt(x+a) + sqrt(x) = a
sqrt(x+a) - sqrt(x) = y

Now multiply the top and bottom equation with each other:

(sqrt(x+a) + sqrt(x))(sqrt(x+a) - sqrt(x)) = ay

Notice that this is just the difference of two squares:

(sqrt(x+a))^2 - sqrt(x)^2 = ay
-> (x+a) - x = ay

The x cancels out, and we neatly have:
a = ay
ay - a = 9
a(y-1) = 0

Thus

Answer: sqrt(x+a) - sqrt(x) = 1 If sqrt(x+a) + sqrt(x) = a for *a* not equal to 0


That’s about it ( I noticed that this is just the third method presented in the video :) Nice! )

[pwmiles]

sqrt(x+a) + sqrt(x) = a
sqrt(x+a) - sqrt(x) = ?

a? = (x + a) - x
a? = a

a=0 or ?=1 (third method I think)

[j.r.]

I used a version of the 3rd method. It's not essential to use u as a new variable; just multiply the second expression by both sides of the first expression, and set the two results equal to each other. The answer pops right out. Anyway, this problem was clearly crying out for multiplication as the simplest approach. It practically said, "Hey, difference of two squares here! Multiply us!"

[geraldillo]

I think it's much easier if you use that fact that √(x+a)-√x-=(√(x+a)+√x)-2√x
and calculate the value of √x and substitute √(x+a)-√x for a

√(x+a)+√x=a
√(x+a)=a-√x
x+a=(a-√x)^2
x+a=a^2-2a√x+x
2a√x=a^2-a
2√x=a-1
√x=(a-1)/2

√(x+a)-√x=(√(x+a)+√x)-2√x
√(x+a)-√x=a-2√x
√(x+a)-√x=a-2((a-1)/2)
√(x+a)-√x=a-(a-1)
√(x+a)-√x=1

[nnwslswu]

I firstly came up with the 3rd method~

[KJIUYHN]

I'm not sure, but I feel something is wrong with first method.
when a is less than 1, √(a-1)^2/4 = (1-a)/2.

[leouser]

It seems that the answer is valid for numbers a greater than or equal to 1 only. There are no solutions for a between 0 and 1.

[chauff]

Set ? = U
We have
a.u= x+a-x=a
If u=0 then a.u= a.0 =0 which is correct with all a
If u≠0 then a.u=x+a-x=a
=>U=1

[walterwen]

Evaluating a Radical Expression in Three Ways: √(x + a) + √x = a; √(x + a) – √x = ?
If: a = 0, x = 0; √(x + a) – √x = 0
√(x + a) + √x = a; a ≠ 0, x > 0
First method:
Let: y = √(x + a) – √x, [√(x + a) + √x][√(x + a) – √x] = ay
(x + a) – x = a = ay, y = 1; √(x + a) – √x = y = 1
Second method:
√(x + a) = a – √x, x + a = a^2 – 2a√x + x, a^2 = 2a√x + a; a = 2√x + 1
√(x + a) + √x = a = 2√x + 1; √(x + a) – √x = 1
Third method:
√x = a – √(x + a), x = [a – √(x + a)]^2 = a^2 – 2a√(x + a) + a + x
a^2 – 2a√(x + a) + a = 0, a = 2√(x + a) – 1
√(x + a) + √x = a = 2√(x + a) – 1; √(x + a) – √x = 1