Question

If a and ß are the zeroes of polynomial x2 + x – 6, Find the value of 1/a+1/ß​

Answer 1

Answer:

$\alpha + \beta = \frac{ - b}{a} = - 1 \\ \alpha \beta = \frac{c}{a} = - 6 \\ now \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \alpha + \beta }{ \alpha \beta } = \frac{ - 1}{ - 6} = \frac{1}{6}$

Answer 2

Given that alpha and beta are the zeroes of polynomial x² + x - 6

we've to find the value of 1/alpha + 1/beta

so let's find the zeroes of the given polynomial first.

using splitting the middle term method,

➡ x² + x - 6 = 0

➡ x² + (3x - 2x) - 6 = 0

➡ x² + 3x - 2x - 6 = 0

➡ x(x + 3) - 2(x + 3) = 0

➡ (x + 3) (x - 2)

➡ x = -3, x = 2

therefore the value of :-

• alpha = -3

• beta = 2

hence, 1/alpha + 1/beta = -1/3 + 1/2

taking LCM of 3 and 2 = 3 × 2 = 6

= (-1 × 2)/(3 × 2) + (1 × 3)/(2 × 3)

= -2/6 + 3/6

= 1/6

the value of 1/α + 1/ß is = 1/6