# If f(x) = x2

Given f(x) = x2 – 3x + 4.

We need to find x satisfying f(x) = f(2x + 1).

We have f(2x + 1) = (2x + 1)2 – 3(2x + 1) + 4

f(2x + 1) = (2x)2 + 2(2x)(1) + 12 – 6x – 3 + 4

f(2x + 1) = 4x2 + 4x + 1 – 6x + 1

f(2x + 1) = 4x2 – 2x + 2

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

x2 – 3x + 4 = 4x2 – 2x + 2

3x2 + x – 2 = 0

3x2 + 3x – 2x – 2 = 0

3x(x + 1) – 2(x + 1) = 0

(x + 1)(3x – 2) = 0

x + 1 = 0 or 3x – 2 = 0

x = –1 or 3x = 2

x = –1 or Thus, the required values of x are –1 and .

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