The zeros of the polynomial x^2 + 1/6x - 2 are (a) -3, 4 ​

Question

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 1

 (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 = \(\frac{4}3\)