Solve the given equation for the real variable x.?

Question

sqrt[x-sqrt(2x-1)]+sqrt[x+sqrt(2x-1)] = sqrt(2).

first term+sqrt[x-sqrt(2x-1)]

sorry it should be +

When I say solve, it means find all solutions.

Many came up only with 1 as a solution.
However, this is not the only real solution.

Answer 1

Square both sides,

2x + 2sqrt(x^2-2x+1) = 2

which can be simplified further into

2x + 2x - 2 = 2, if x is greater than 1

x = 1

or 2x - 2x + 2 = 2,if x is less than 1

which leads to x can be any number between 1/2 and 1.
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Puggy, I read your solution. Please read my solution carefully. Did you notice what you missed?

Answer 2

So you want to solve

sqrt [(x - sqrt(2x - 1)] + sqrt [x + sqrt(2x - 1)] = sqrt(2)

One of the best first steps would be to multiply the left hand side and right hand side by the conjugate of what you see on the left hand side. The reason why is because we will get rid of those outer square root symbols on the left hand side all in one shot. Recall that that conjugate of (a + b) is (a - b), and multiplying by the conjugate yields the result a^2 - b^2.

In our case, we'll have sqrt(junk) + sqrt(junk) turn into (junk)^2 - (junk)^2, WITHOUT any square root symbols.

Multiplying both sides by sqrt [(x - sqrt(2x - 1)] - sqrt [x + sqrt(2x - 1)], we get

[x - sqrt(2x - 1)] - [x + sqrt(2x - 1)] =
sqrt(2) { sqrt [(x - sqrt(2x - 1)] - sqrt [x + sqrt(2x - 1)] }

(Going to emphasize again how the square root symbol has disappeared on the left hand side). Now, we can simplify the left hand side.

x - sqrt(2x - 1) - x - sqrt(2x - 1) =
sqrt(2) { sqrt [(x - sqrt(2x - 1)] - sqrt [x + sqrt(2x - 1)] }

And the x terms on the left hand side actually cancel out, while the square roots merge.

-2sqrt(2x - 1) =
sqrt(2) { sqrt [(x - sqrt(2x - 1)] - sqrt [x + sqrt(2x - 1)] }

If we divide both sides by the sqrt(2), then -2/sqrt(2) would be the same as -2/2^(1/2), or -(2^1)/2^(1/2) = -2^(1/2), or -sqrt(2). So we can adjust the left hand side accordingly to

-sqrt(2) * sqrt(2x - 1) =
sqrt [(x - sqrt(2x - 1)] - sqrt [x + sqrt(2x - 1)]

At this point, we have no choice but to square both sides. Remember that on the left hand side, we square EACH term, since it's a product. On the right hand side, it's going to look VERY messy but nevertheless, we perform FOIL on it. We should obtain

2(2x - 1) =
[(x - sqrt(2x - 1)] - 2 sqrt [(x - sqrt(2x - 1)] sqrt [x + sqrt(2x - 1)]
+ [x + sqrt(2x - 1)]

Note that the first term and last term on the right hand side are mergeable. Also, the -sqrt(2x -1) cancels with the +sqrt(2x - 1), to get;

2(2x - 1) = 2x - 2 sqrt [(x - sqrt(2x - 1)] sqrt [x + sqrt(2x - 1)]

Remember that whenever we are multiplying two square roots, we can multiply their insides (i.e. sqrt(z) * sqrt(a) = sqrt(za)). In our case though, we're apparently multiplying together two conjugates, which would eliminate the INNER square roots, and inside the square root, we would again get [junk]^2 - [junk]^2.

2(2x - 1) = 2x - 2 sqrt [x^2 - (2x - 1)]

Whew! Now, let's simplify the left hand side, and then move the 2x on the right hand side to the left hand side.

4x - 2 - 2x = - 2 sqrt [x^2 - (2x - 1)]
2x - 2 = - 2 sqrt [x^2 - 2x + 1]

Dividing everything by 2, we get

x - 1 = - sqrt [x^2 - 2x + 1]

Turns out that x^2 - 2x + 1 is a square binomial; it's in fact
(x - 1)^2. Therefore,

x - 1 = - sqrt ( [x - 1]^2 )

The square root of a square number is itself.

x - 1 = - (x - 1)

Solve as normal.

x - 1 = -x + 1
2x = 2
x = 1

Normally, when solving equations with radicals, we should test our solution.

Let x = 1. Then
LHS = sqrt [(1 - sqrt(2(1) - 1)] + sqrt [1 + sqrt(2(1) - 1)]
= sqrt [(1 - sqrt(1))] + sqrt [1 + sqrt(1)]
= sqrt [0] + sqrt(2)
= 0 + sqrt(2)
= sqrt(2) = RHS

Therefore, this solution checks out, and our solution is
x = 1.

Answer 3

Solution :

Sqrt[x-sqrt(2x-1)]+sqrt[(x+sqrt(2x-1)]=sqrt(2)

Step 1: Square the equation both side

x-sqrt(2x-1)+x+sqrt(2x-1)=2

Step 2 : Squaring the Equation again both side
X*2-(2x-1)+X*2+(2x-1)=4

2x*2=4
X*2=4/2
X=Sqrt(2) - Answer proved