Nice! I first brute-forced by successive squaring; then thought to make the substitution x = sin^2 theta. As it turns out, that substitution is the identical algebra as your first solution, but with more characters to write on each line, and cos theta appearing instead of your z and y.
It certainly has a Pythagoras feel to it. Also substituting X=cos^2 theta leaves the expression looking like a quadratic formula!!! Unfortunately neither of these identities help to find a short cut to the answer.
I substituted y^2 = x giving me y + √(y^2 - √(1-y^2)) = 1 y^2 - √(1-y^2) = (1-y)^2 Move things around and we have difference of squares y^2 - (1-y)^2 = √(1-y^2) The LHS reduces to 2y-1 so (2y-1)^2 = 1-y^2 The 1s cancel as does the y (y /= 0) and we have y=4/5 So x = y^2 = 16/25 5y = 4
Let A=sqrt(x) B=sqrt(1-x) so B^2 = 1- A^2 or A^2 + B^2 =1 and the original equation becomes A + sqrt(A^2 -B)=1 Thus 1-A= sqrt(A^2 -B) square this to get (1-A)^2 =1 -2A + A^2 =A^2 - B or B=2A-1 substitute into A^2 + B^2 =1 A^2 + (2A-1)^2 =1 or 5(A^2) -4A =0 or A*(5A-4) =0 A=0 leads to an invalid result so A=4/5 and x=16/25.
Pretty simp to be honest. The only solution is 16/25. Just keep squaring and canceling and you'll end up with two solutions, the other is x=0 which is extraneous
Nice!
I first brute-forced by successive squaring; then thought to make the substitution x = sin^2 theta. As it turns out, that substitution is the identical algebra as your first solution, but with more characters to write on each line, and cos theta appearing instead of your z and y.
Nice work!
It certainly has a Pythagoras feel to it. Also substituting X=cos^2 theta leaves the expression looking like a quadratic formula!!! Unfortunately neither of these identities help to find a short cut to the answer.
Very interesting first method.
Glad you think so!
Excelentes métodos
I substituted y^2 = x giving me
y + √(y^2 - √(1-y^2)) = 1
y^2 - √(1-y^2) = (1-y)^2
Move things around and we have difference of squares
y^2 - (1-y)^2 = √(1-y^2)
The LHS reduces to 2y-1 so
(2y-1)^2 = 1-y^2
The 1s cancel as does the y (y /= 0) and we have y=4/5
So x = y^2 = 16/25
5y = 4
Let A=sqrt(x) B=sqrt(1-x) so B^2 = 1- A^2 or A^2 + B^2 =1 and the original equation becomes A + sqrt(A^2 -B)=1
Thus 1-A= sqrt(A^2 -B) square this to get (1-A)^2 =1 -2A + A^2 =A^2 - B or B=2A-1 substitute into A^2 + B^2 =1
A^2 + (2A-1)^2 =1 or 5(A^2) -4A =0 or A*(5A-4) =0 A=0 leads to an invalid result so A=4/5 and x=16/25.
Nice and easy
I wouldn't say easy. Not unless you are familiar with these types of problems. I had to try a few different things before I just squared it anyway!
Pretty simp to be honest. The only solution is 16/25.
Just keep squaring and canceling and you'll end up with two solutions, the other is x=0 which is extraneous