Question

If α andβ are the zeroes of the equation 6x2+x-2
find 1/α + 1/β

Answer 1

⇒ 6x² + x - 2

⇒ 6x² - 3x + 4x - 2

⇒ 3x (2x - 1) + 2 (2x - 1)

⇒ (3x + 2) (2x - 1)

∴ Zeroes are -2/3 and 1/2.

Let   α = -2/3   &   β = 1/2  

⇒  1/α + 1/β

⇒  -3/2  +  2/1

⇒  -3/2  +  4/2

⇒  (- 3 + 4) / 2

⇒  1/2

∴  Answer is 1/2.

Answer 2

Hey,

$f(x) = 6 {x}^{2} + x - 2 \\ \: \: \: \: \: \: \: \: \: = 6x {}^{2} + 4x - 3x - 2 \\ \: \: \: \: \: \: \: \: \: \: = (3x + 2)(2x - 1) \\ \\ \\ f(x) = 0 \\ (3x + 2)(2x - 1) = 0 \\ x = - \frac{ - 2}{3} \: or \: \frac{1}{2 } \\ \alpha = \frac{ - 2}{3} \: and \: \beta = \frac{1}{2} \\ \\ now \\ \alpha + \beta = \frac{ - 2}{3} + \frac{1}{2} = \frac{ - 1}{6} \\ \\ \alpha \beta = \frac{ - 2}{3} \times \frac{1}{2} = \frac{ - 1}{3} \\ \\ \\ hence \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \alpha + \beta }{ \alpha \beta } = \frac{ \frac{ - 1}{6} }{ \frac{ - 1}{3} } = \frac{ - 1}{6} \times \frac{ - 3}{1} = \frac{3}{6} = \frac{1}{2}$

I hope it helps you!!