Question

If alpha and beta are the zeroes of the polynomial p (t)= 6t^2+t+12,then evaluate alpha + beta?

Answer 1

Answer:

- 1/6

Step-by-step explanation:

Given : 6t² + t - 12

By Middle Term Factorisation, we get

→ 6t² + 9t - 8t - 12

→ 3t(2t + 3) - 4(2t + 3)

→ (3t - 4)(2t + 3)

To find the zeroes, these factors should be equal to 0.

∴ 3t - 4 = 0 and 2t + 3 = 0

3t = 4 and 2t = - 3

t = 4/3 and t = - 3/2

Given that α and β are the zeroes of the above polynomial, so

α = 4/3 and β = - 3/2

Now,

α + β = (4/3) + (- 3/2)

= (4/3) - (3/2)

= [4(2) - 3(3)] / 6

= (8 - 9)/6

= - 1/6

Answer 2

Step-by-step explanation:

Given,

$p(t) = 6 {t}^{2} + t + 12$

Here,

$a = 6 \\ b = 1 \\ c = 12$

To find

$\alpha + \beta$

As we know ,

$\alpha + \beta = \frac{ - b}{a}$

so put the values,

$\alpha + \beta = \frac{ - (1)}{6}$

Hence ,the value is -1/6