Question

If f(x)=x^2 the find the value of x for which f(x=f(2x+1)

Answer 1

Step-by-step explanation:

$f(x) = {x}^{2} \\ f(2x + 1) = (2x + 1) ^{2} \\ = 4 {x}^{2} + 4x + 1 \\ = 4 {x}^{2} + 2x + 2x + 1 \\ = 2x(2x + 1) + 1(2x + 1) \\ = (2x + 1)(2x + 1) \\ 2x = - 1 \\ x = \frac{ - 1}{2} and \frac{ - 1}{2}$

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