Question:
If f(x) = x2 − 3x + 4, then find the values of x satisfying the equation f(x) = f(2x + 1).
Solution:
Given:
f (x) = x2 – 3x + 4
Therefore,
f (2x + 1) = (2x + 1)2 – 3(2x + 1) + 4
= 4x2 + 1 + 4x – 6x – 3 + 4
= 4x2 – 2x + 2
Now,
f (x) = f (2x + 1)
⇒ x2 – 3x + 4 = 4x2 – 2x + 2
⇒ 4x2 – x2 – 2x + 3x + 2 – 4 = 0
⇒ 3x2 + x – 2 = 0
⇒ 3x2 + 3x – 2x – 2 = 0
⇒ 3x(x + 1) – 2(x +1) = 0
⇒ (3x – 2)(x +1) = 0
⇒ (x + 1) = 0 or ( 3x – 2) = 0
$\Rightarrow x=-1$ or $x=\frac{2}{3}$
Hence, $x=-1, \frac{2}{3}$.