Question

If f(x) = x² - 3x + 4, then find the value of x satisfying f(x) = f(2x + 1).

Answer 1

f(x) = x² - 3x + 4

f(2x + 1) = (2x + 1)² - 3(2x + 1) + 4

= 4x² + 4x + 1 - 6x - 3 + 4

= 4x² + 2x + 2


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

x² - 3x + 4 = 4x² + 2x + 2

or, 4x² - x² + 2x + 3x + 2 - 4 = 0

or, 3x² + 5x - 2 = 0

or, 3x² + 6x - x - 2 = 0

or, 3x(x + 2) - 1(x + 2) = 0

or, (3x - 1)(x + 2) = 0

hence, x = -2 and 1/3

Answer 2

Answer:

-1 and 2/3

Step-by-step explanation:

f(x) = x² - 3x + 4

f(2x + 1) = f(t) = t² - 3t + 4, where t = 2x+1

But given:

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

x² - 3x + 4 = t² - 3t + 4

x² - 3x = t² - 3t

t² -x² - 3t + 3x = 0

(t - x)(t+x) -3(t - x) = 0

(t - x) [t + x - 3] = 0

(2x +1 - x) [2x +1 + x - 3] = 0

(x +1) [3x - 2] = 0

x + 1 = 0, 3x - 2 =0

x = -1, 2/3

The required values of x are

-1 and 2/3