If $f (x) = x^{2} - 3 x + 4$ and $f (x) = f (2 x + 1)$ , the value of x is

-1 or 2/3

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1 or-2/3

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-1 or 3/2

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1 or -3/2

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Solution

The correct option is A -1 or 2/3
From given, we have,

$f (x) = f (2 x + 1)$

$x^{2} - 3 x + 4 = {(2 x + 1)}^{2} - 3 (2 x + 1) + 4$

$x^{2} - 3 x + 4 = 4 x^{2} + 4 x + 1 - 6 x - 3 + 4$

$x^{2} - 3 x + 4 = 4 x^{2} - 2 x + 2$

$- 3 x^{2} - x + 2 = 0$

$x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (- 3) 2}}{2 (- 3)}$

$x_{1} = \frac{- (- 1) + \sqrt{{(- 1)}^{2} - 4 (- 3) 2}}{2 (- 3)} = - 1$

$x_{2} = \frac{- (- 1) - \sqrt{{(- 1)}^{2} - 4 (- 3) 2}}{2 (- 3)} = \frac{2}{3}$

$∴ x = - 1, \frac{2}{3}$

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