Given:
f(x) = x2 – 3x + 4.
Let us find x satisfying f (x) = f (2x + 1).
We have,
f (2x + 1) = (2x + 1)2 – 3(2x + 1) + 4
= (2x) 2 + 2(2x) (1) + 12 – 6x – 3 + 4
= 4x2 + 4x + 1 – 6x + 1
= 4x2 – 2x + 2
Now, f (x) = f (2x + 1)
x2 – 3x + 4 = 4x2 – 2x + 2
4x2 – 2x + 2 – x2 + 3x – 4 = 0
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 \(\dfrac{2}{3}\)
The values of x are –1 and \( \dfrac{2}{3}\)
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