Question

02. Without solving, find the sum and product of zeroes:
(a)
2(x² - 1) + 3x - 9
​

Answer 1

Given Polynomial

2(x² - 1) + 3x - 9

First of all simplify this :-

→ 2x² - 2 + 3x - 9

→ 2x² + 3x - 9 - 2

→ 2x² + 3x - 11

Now, if the zeroes of the given Polynomial are $\sf\alpha \ and \ \beta$, then we know that :-

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

and, $\sf \alpha\beta = \frac{c}{a}$

Here, comparing with the form ax² + bx + c, we get

→ a = 2

→ b = 3

→ -b = -3

→ c = - 11

So, sum of zeroes =

$\sf \alpha + \beta = \frac{-b}{a} \\ \\ \implies \alpha + \beta= \frac{-3}{2}$

and, product of zeroes =

$\sf \alpha\beta = \frac{c}{a} \\ \\ \implies \alpha\beta = \frac{-11}{2}$

Hence, the answer is :-

$\sf \frac{-3}{2} \ and \ \frac{-11}{2}$