Solve \(f(x)=f(2x+1)\)

Solution

Given: $$ f(x)=x^2-3x+4 $$

First find \(f(2x+1)\):

$$ f(2x+1)=(2x+1)^2-3(2x+1)+4 $$

$$ =4x^2+4x+1-6x-3+4 $$

$$ =4x^2-2x+2 $$

Now, $$ f(x)=f(2x+1) $$

$$ x^2-3x+4=4x^2-2x+2 $$

$$ 3x^2+x-2=0 $$

$$ 3x^2+3x-2x-2=0 $$

$$ 3x(x+1)-2(x+1)=0 $$

$$ (x+1)(3x-2)=0 $$

Therefore, $$ x=-1 $$ or $$ x=\frac{2}{3} $$

Hence, $$ \boxed{x=-1,\ \frac{2}{3}} $$