Hi having a bit of a mind blank trying to find the solutions to this quadratic equation, any hints would be invaluable
x^2 -(5/3)x +4/3=0
The question is not clear.. What does fixed point of an equation mean?Fixed points are defined for functions (as far as I know).
Also usually when one says "solve an equation", one should specify the domain... Is x real?
So assuming we want x real, is the question asking for the fixed point of the function f(x) = x^2 -(5/3)x +4/3 ?
You are asked to find fixed points of f(x) = x^2 -(5/3)x +4/3. Read this, to know the definition.
Now we know that a solution to f(x) = x is called a fixed point of f(x).
So find values of x for which f(x) = x.
f(x) = x implies x^2 -(5/3)x +4/3 = x, which implies x^2 -(8/3)x +4/3=0
Now hit it with the quadratic formula and get x. Do you get it now?
Just basic algebraic manipulations:
$\displaystyle x^2 - \frac53 x + \frac43 = x \implies \left(x^2 - \frac53 x + \frac43 \right) - x = 0 \implies x^2 - \left(\frac53 + 1 \right) x + \frac43$ $\displaystyle = 0 \implies x^2 - \frac83 x + \frac43 = 0
$
The only thing I have not explained is $\displaystyle 1 + \frac53 = \frac83$.