Solving A Radical Equation | Algebra #challengingmathproblems

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[plutlorian]

I learn a lot from you, you are great

[MichaelRothwell]

Here is my solution.

To solve √(1-x)=1-x²

By inspection, x=0 and x=1 are solutions.
Also, for LHS to be defined x≤1, and, as LHS≥0, we have x²≤1, so |x|≤1 (which implies the first condition x≤1).

Squaring both sides,
1-x=(1-x²)² (but we must bear in mind that |x|≤1)
1-x=(1-x)(1+x)(1-x²)
(1-x)(1-(1+x)(1-x²))=0
(1-x)(1-(1+x-x²-x³))=0
(1-x)(-x+x²+x³)=0
(1-x)x(x²+x-1)=0
So x=0 or x=1 or x²+x-1=0

Now the equation x²-x-1=0 has roots φ (the golden ratio) & -1/φ, and the roots of x²+x-1=0 are the opposites (replace x by -x in either equation to get the other), so are -φ & 1/φ.
However, -φ<-1, so must be rejected.

So the solutions to our equation are x=0, 1/φ, 1.

[laurendoe]

I don't understand where the step "x^2 - y^2 = x - y" came from.

[AlexD]

How did he get y=x like that towards the end?

[gdmathguy]

Idea: Take a stupid "difficult" problem from social media, solve it amd make the solution into a video that's similar to a video McNally would make of a bad lock