Question:
If $f(x)=x^{2}-3 x+4$ and $f(x)=f(2 x+1)$, find the values of $x$.
Solution:
Given: $f(x)=x^{2}-3 x+4$ (1)
and f(x) = f(2x + 1)
Need to Find: Value of x
Replacing x by (2x + 1) in equation (1) we get,
$f(2 x+1)=(2 x+1)^{2}-3(2 x+1)+4 \cdots(2)$
According to the given problem, f(x) = f(2x + 1)
Comparing (1) and (2) we get,
$x^{2}-3 x+4=(2 x+1)^{2}-3(2 x+1)+4$
$\Rightarrow x^{2}-3 x+4=4 x^{2}+4 x+1-6 x-3+4$
$\Rightarrow 4 x^{2}+4 x+1-6 x-3+4-x^{2}+3 x-4=0$
$\Rightarrow 3 x^{2}+x-2=0$
$\Rightarrow 3 x^{2}+3 x-2 x-2=0$
$\Rightarrow 3 x(x+1)-2(x+1)=0$
$\Rightarrow(3 x-2)(x+1)=0$
So, either $(3 x-2)=0$ or $(x+1)=0$
Therefore, the value of $x$ is either $\frac{2}{3}$ or $-1$ [Answer]