If f(x) = x2 – 3x + 4, then find the values of x satisfying the equa

1.	If f(x) = x2 – 3x + 4, then find the values of x satisfying the equation f(x) = f(2x + 1).
Answer» 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 \(\frac{2}{3}\) Thus, The required values of x are –1 and \(\frac{2}{3}\).

If f(x) = x2 – 3x + 4, then find the values of x satisfying the equation f(x) = f(2x + 1).

Answer»

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 \(\frac{2}{3}\)

Thus,

The required values of x are –1 and \(\frac{2}{3}\).