Solve this

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]