Question

if alfa & beta are the zeroes of x2 + 5x + 6 find the value of alfa-1 + beta-1​

Answer 1

Solution :

Given :

$\mathsf { P(x) \: = \: x^2 \: + \: 5x \: + \: 6} \\ \sf\alpha \: and \: \beta \: are \: zeroes \: of \: P(x).$

Here :

$\sf Coefficient \: of \: x^2 (a) \: = \: 1 \\ \sf Coefficient \: of \: x (b) \: = \: 5 \\ \sf Constant \: term (c) \: = \: 6$

Now :

$\mathsf{\implies Sum \: of \: zeroes \: = \: \dfrac{-b}{a}} \\ \mathsf{\implies \alpha \: + \: \beta \: = \: \dfrac{-5}{1} } \\ \mathsf{ \: \: \therefore \: \: \alpha \: + \: \beta \: = \: -5 \: \: \: \: \: \: \: \: ...(1)} \:$

And :

$\mathsf{ \implies Product \: of \: zeroes \: = \: \dfrac{c}{a}} \\ \mathsf{ \implies \alpha \: \times \: \beta \: = \: \dfrac{6}{1}} \\ \mathsf{ \: \: \therefore \: \: \alpha\beta \: = \: 6 \: \: \: \: ...(2)} \:$

To Find :

$\mathsf{ = \: {\alpha}^{-1} \: + \: {\beta}^{-1}} \\ \mathsf{ = \: \dfrac{1}{\alpha} \: + \: \dfrac{1}{\beta}} \\ \mathsf{ = \: \dfrac{ \alpha \: + \: \beta}{\alpha\beta}} \\ \\ \sf Put \: the \: value \: of \: (1) \: and \: (2), \: \\ \mathsf{ = \: \dfrac{ - 5}{6} }$

$\mathfrak {The \: required \: answer \: is \: \dfrac{-5}{6}}.$

Answer 2

Nai samajh aya to koi baat nahi . . . .

Vse bhi bs time pàss hota tha ¯\_ಠ_ಠ_/¯

Purane vérsion me.

Lekin bada maza ata tha ಠ◡ಠ