The zeros of the polynomial x2 + \(\frac{1}6\)x - 2 are(a) -3, 4 (b

1.	The zeros of the polynomial x2 + \(\frac{1}6\)x - 2 are(a) -3, 4 (b) \(\frac{-3}{2},\frac{4}3\)(c) \(\frac{-4}{3},\frac{3}2\)(d) none of these
Answer» (b) \(\frac{4}{3},\frac{-3}2\) Let f(x) = x² + \(\frac{1}6\) x – 2 = 0 ⇒ 6x² + x – 12 = 0 ⇒ 6x² + 9x – 8x – 12 = 0 ⇒ 3x (2x + 3) –4 (2x + 3) = 0 ⇒ (2x + 3) (3x – 4) = 0 ∴x = \(\frac{-3}2\) or x = \(\frac{4}3\)

The zeros of the polynomial x2 + \(\frac{1}6\)x - 2 are(a) -3, 4 (b) \(\frac{-3}{2},\frac{4}3\)(c) \(\frac{-4}{3},\frac{3}2\)(d) none of these

Answer»

(b) \(\frac{4}{3},\frac{-3}2\)

Let f(x) = x² + \(\frac{1}6\) x – 2 = 0

⇒ 6x² + x – 12 = 0

⇒ 6x² + 9x – 8x – 12 = 0

⇒ 3x (2x + 3) –4 (2x + 3) = 0

⇒ (2x + 3) (3x – 4) = 0

∴x = \(\frac{-3}2\) or x = \(\frac{4}3\)