A Sum of Roots of Roots

Square both sides to get $$x + 2\sqrt{\left(x + \sqrt{2x - 1}\right)\left(x - \sqrt{2x - 1}\right)} + x = 100$$ $$2x + 2\sqrt{x^2 - 2x + 1} = 100$$ $$x + \sqrt{(x - 1)^2} = 50$$ $$x + |x - 1| = 50$$ If $x - 1 \leq 0$ , then we get $1 = 50$ , so no solution.

If $x - 1 \geq 0$ , then we get $$x = \frac{51}{2}$$ Hence $$a + b = 51 + 2 = \boxed{53}$$

This is basically second problem from IMO 1959.

Squaring both sides gives $x + \sqrt{2x - 1} + x - \sqrt{2x - 1} + 2\sqrt{x^2 - 2x + 1} = 100$ which can be simplified to $2x + 2\sqrt{(x - 1)^2} = 100$ which can further be simplified to $4x - 2 = 100$ solving for x we get $\dfrac{51}{2}$ , 51 + 2 = 53, our final answer.

To get rid of the nested radicals, let's both sides of the equation. This gives us $x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{x^2-(2x-1)}=100$ Cancelling out and factoring gives us $2x+2\sqrt{(x-1)^2}=100$ $2x+2(x-1)=100$ $2x-1=50$ $2x=51$ $x=\frac{51}{2}$ $51+2=\boxed{53}$

Let a=x, b= $\sqrt{2x-1}$

$\sqrt{a+b}$ + $\sqrt{a-b}$ =10

Square both sides

a+b+2 $\sqrt{(a+b)(a-b)}$ +a-b=100

Rearranging and divide by 2 gives

$\sqrt{(a+b)(a-b)}$ =50-a

Squaring both sides gives :

$a^{2}$ - $b^{2}$ =2500-100a+ $a^{2}$

Substitute $b^{2}$ =2x-1

And a=x

$a^{2}$ -2a+1=2500-100a+ $a^{2}$

Rearranging gives:

98a=2499

a= $\frac{51}{2}$

51+2=53

\sqrt{x + \sqrt{2x - 1}} + \sqrt{x - \sqrt{2x - 1}} = 10 \Rightarrow (\sqrt{x + \sqrt{2x - 1}} + \sqrt{x - \sqrt{2x - 1}}) ^ 2 = 100 \Rightarrow 2x + 2 \times (\sqrt{x + \sqrt{2x - 1}} \times \sqrt{x - \sqrt{2x - 1}}) = 100 \Rightarrow 2x+ 2 \times ( x - 1) = 100 \Rightarrow 4x = 102 \Rightarrow x = \frac {51}{2} Therefore a + b = 51 + 2 = 53

$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=10$ $=>(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}})^2=10^{2}$ $=>(\sqrt{x+\sqrt{2x-1}})^2+2*\sqrt{x+\sqrt{2x-1}}*\sqrt{x-\sqrt{2x-1}}+(\sqrt{x-\sqrt{2x-1}})^{2}=100$ $=>x+\sqrt{2x-1}+2*\sqrt{(x+\sqrt{2x-1})( {x-\sqrt{2x-1} )}}+x-\sqrt{2x-1}=100$ $=>2x+2*\sqrt{x^{2}-(\sqrt{2x-1})^{2}}=100$ $=>2x+\sqrt{x^{2}-2x+1}=100$ $=>2x+2*(x-1)=100$ $=>2x+2x-2=100$ $=>4x=102$ $=>x=\frac{ 102 }{ 4 }=\frac{ 51 }{ 2 }$ Therefore,a=51,b=2 a+b=53

square both sides to get $(x + \sqrt{2x-1}) + 2\sqrt{(x + \sqrt{2x-1})(x-sqrt{2x-1})} + (x - \sqrt{2x-1}) = 100$ the first and third terms can be added for simplification, and the second term can also be simplified using difference of two squares. the result is this: $2x + 2\sqrt{x^2-2x+1}=100$ the second term involves the square root of a perfect square, so it becomes: $4x-2=100$ this is easily solved as: $x=\frac{51}{2}$ so the solution is $51+2=53$

lets (2x-1)=t^2 so we have x= (t^2+1)/2 => sqrt(t^2+1+2t)/2 + sqrt(t^2+1-2t)/2 = 10 => solving the Bracket => t+1 + t-1 = 10sqrt2 => t=5sqrt2 => t^2 = 50 => x = 51/2 = a/b so a+b= 53 :D

squaring on both side we get

x+√2x+1 +x-√2x-1 +2√(x-1)2 =100

2x+2(x-1)=100

x+x-1=50

2x=51

x=51/2

where a=51 and b=2

so a+b= 51+2=53

First off, we shift over one of the square roots.

$\sqrt{x + \sqrt{2x - 1}} = 10 - \sqrt{x - \sqrt{2x -1 }}$

Squaring each side, we arrive at:

$x + \sqrt{2x - 1} = 100 - 20\sqrt{x - \sqrt{2x - 1}} + x - \sqrt{2x - 1}$

Continuing,

$20\sqrt{x - \sqrt{2x - 1}} = 100 - 2\sqrt{2x - 1}$ $10\sqrt{x - \sqrt{2x - 1}} = 50 - \sqrt{2x - 1}$

Square again,

$100x - 100\sqrt{2x - 1} = 2500 - 100\sqrt{2x -1} + 2x - 1$

$100x = 2500 + 2x - 1$

$98x = 2499$

$x = \frac{2499}{98} = \frac{51}{2}$

Thus $a + b = 51 + 2 = 53$

$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=10$ $\rightarrow$ Square both sides, we get $x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}=100$ $\Leftrightarrow$ $2x+2\sqrt{x^2-2x+1}=100$ $\Leftrightarrow$ $2x+2\sqrt{(x-1^2)}=100$ $\Leftrightarrow$ $2x+2(x-1)=100$ $\Leftrightarrow$ $4x=102$ $\Leftrightarrow$ $x=\frac{102}{4}=\frac{51}{2}$ So, $a+b=51+2=53$

√(x+√(2x-1)) +√(x-√(2x-1)) = 10

〖(√(x+√(2x-1)) +√(x-√(2x-1)) )〗^2= 〖10〗^2

x + √(2x-1) + 2.√((x+√(2x-1)).(x-√(2x-1)) ) + x - √(2x-1) = 100

2x + 2(√(x^2-2x+1)) = 100

2x + 2(x-1) = 100

x + x-1 = 50

2x = 51

x = 51/2

a = 51, b = 2

a+b = 51 + 2 = 53

Squaring both sides of the equation gives

2x + 2root(x^2-2x+1)=100

2x+2(x-1)=100

4x=102

x=51/2

Therefore 51+2 =53

Squaring both sides we get $x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}=10^2$ $2x+2\sqrt{x^2-2x+1}=100$ $\Rightarrow$ $2x+2(x-1)=100$ $\Rightarrow$ $4x=102$ $\Rightarrow$ $x=\frac {51}{2}$ $\Rightarrow$ $a+b=53$

Let (2x-1)^1/2=y =>2x-1=y^2; by putting y in the given equation... (x+y)^1/2 + (x-y)^1/2 =10 taking both side square, 2x+2*(x^2-y^2)^1/2=100; by putting value of y^2=2x-1 in the equation 2x+2(x-1)=100; =>x=51/2=a/b; a+b=53 sol.

Pretty simple algebra, anybody who knows how to square and knows the factors of two perfect squares can do this easily

1. Square both sides: $100 = x + /sqrt{2x-1} + x - /sqrt{2x-1} + 2/sqrt{x + /sqrt{2x-1}}/sqrt{x - /sqrt{2x-1}}$
2. Simplify: $100 = 2x +2\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}$
3. Divide each side by two, and multiply inside the radical : $50 = x + \sqrt{x^{2}-2x+1}$
4. Factor trinomial: $50 = x + \sqrt{(x-1)^{2}}$
5. Simplify: $50 = x + x - 1$
6. Simplify: $51 = 2x$
7. Simplify: $x = 51/2 = a/b$
8. Solve: $a+b = 51+2 = 53$

If we square the entire expression, we get $2x+2\sqrt{x^2-2x+1}=2x+2\sqrt{(x-1)^2}=100$ , from which the answer immediately follows.

Let $ a=\sqrt{x+\sqrt{2x-1}} , b=\sqrt{x-\sqrt{2x-1}} $ We have: $m+n=10$ $mn=|x-1|$ $m^2 + n^2 =2x$ So, $2x+2|x-1|= (m+n)^2 = 10^2 =100$ So, $x=\frac{51}{2} So, a+b=53

Square both sides to get:

$2x + 2\sqrt{x^{2}-2x+1} = 100$

$2x + 2(x-1) = 100$

$4x = 102$

Thus, $x = 51/2$ , so our answer is 53

sqrt( x + sqrt(2x - 1) ) + sqrt( x - sqrt(2x - 1) ) = 10

sqrt( x + sqrt(2x - 1) ) = 10 - sqrt( x - sqrt(2x - 1) ) "by squaring both sides"

x + sqrt(2x -1) = 100 - 20*sqrt( x - sqrt(2x - 1) ) + x - sqrt(2x -1)

2 sqrt(2x -1) = 100 - 20 sqrt( x - sqrt(2x - 1) )

sqrt(2x -1) = 50 - 10*sqrt( x - sqrt(2x - 1) )

> 10*sqrt( x - sqrt(2x - 1) ) = 50 - sqrt(2x -1) "by squaring both sides"

100 ( x - sqrt(2x - 1) ) = 2500 - 100 sqrt(2x - 1) + 2x - 1

100x - 100 sqrt(2x - 1) = 2500 - 100 sqrt(2x - 1) + 2x - 1

> 98x = 2499

>>>> x = 2499/98 = 51/2 = a/b

then a+b = 53

$(\sqrt{x + \sqrt{2x-1}} +\sqrt{x + \sqrt{2x-1}}) ^ 2 = 4x - 2 =100$

$x = 51/2$

Answer is 51 + 2 = 53

Square both sides, and simplify,then we get the following expression: 2(x^2-(2x-1))^0.5 + 2x = 100, then (x-1) + x = 50, So x=51/2 = a/b. So a+b = 53