Question

\(f(x) = x^2 + \pi x + 1\) and \(\alpha, \beta\) are zeroes of \(f(x)\) find the value of \((\alpha + 1)(\beta + 1)\)

Answer 1

@Jhannybean here is a challenging question that can be tried using vietas formula

Answer 2

what is "vietas formula"?

Answer 3

Oh so first you would do the product rule

Answer 4

its Aaron @Nnesha and not aaroon.

Answer 5

\[
(x-\beta)(x-\alpha)=x^2-\alpha x-\beta x+\alpha\beta=x^2+(-\alpha-\beta)x+\alpha\beta
\]

Answer 6

^^

Answer 7

Bleh, confused.

Answer 8

@AaronAndyson
wio's reply pretty much summarizes the vieta's formula, just compare the coefficients both sides of equation in wio's reply

Answer 9

Ohh, wait wait wait

Answer 10

@Nnesha it's okay, nest time kindly take care.

Answer 11

\[x^2+(-\alpha-\beta)x+\alpha\beta
\]Corresponds to \[
x^2+\pi x+1
\]

Answer 12

Dangit wio.

Answer 13

thanks @ganeshie8

Answer 14

Yeahh that was one of the forms wasn't it.

Answer 15

Then remember that: \[
(\alpha+1)(\beta+1)= \alpha\beta + \alpha+\beta+1
\]I don't even know what the formula is.

Answer 16

Does it make sense now @Jhannybean ?

Answer 17

Kind of,

Answer 18

So what is the answer?

Answer 19

-.-

Answer 20

\(\color{blue}{\text{Originally Posted by}}\) @wio
\[x^2+(-\alpha-\beta)x+\alpha\beta
\]Corresponds to \[
x^2+\pi x+1
\]
\(\color{blue}{\text{End of Quote}}\)

Trying to understand this atm.

Answer 21

compare apples to apples
and coconuts to coconuts

its an art

Answer 22

Think of them as being dot products with the vector \(\langle x^2,x,1\rangle\).

Answer 23

\[x^2+\color{red}{(-\alpha-\beta)}x+\color{blue}{\alpha\beta} \]

Ccprresponds to

\[x^2+\color{red}{\pi} x+\color{blue}{1}\]

compare the color coded parts

Answer 24

Can you elaborate a little more on how I to compare them?

Answer 25

to*

Answer 26

hmm that is quadratic equation
use headphone method
find two number if you multiply them you should get 1
and if you add or subtract them you should get \[\pi \] :P
:D that's all i know

Answer 27

i got this

Answer 28

yes @PFEH.1999 the answer should look something simpler like that.. il confirm one sec..

Answer 29

For a quadratic equation of the form \[y(x) =ax^2+bx+c\]
With roots \(\alpha,\beta\)

the sum of the roots (zeroes):\[\alpha+\beta=-b/a\] and the product of the roots:\[\alpha\beta=c/a\]

For the quadratic
\[f(x)=x^2+\pi x+1\]\[a=1,\quad b=\pi,\quad c=1\]\[\Downarrow\]\[\alpha+\beta = -\pi,\qquad\alpha\beta=1\]

\[(\alpha+1)(\beta+1)= \alpha\beta + (\alpha+\beta)+1 \\
\qquad\qquad=1-\pi+1 \\\qquad\qquad= 2-\pi\]

Answer 30

yes i used @UnkleRhaukus method fo alpha + beta and alpha.beta...i forgot that alpha + beta = -b/a i used -b/2a :D

Answer 31

\[\Uparrow\]

Answer 32

Ah i see how you got fractions ;) @Jhannybean does Unkles answer make sense more or less..

Answer 33

@UnkleRhaukus ' explanation makes things easier to understand.

It's like Vieta's Formula for dummies.

Answer 34

you're right "vieta's formula" is just a fancy name for that method ;p

Answer 35

Sorry I'm so slow at this... still learning

Answer 36

Its all good :D Next problem going to be a bit more challenging !

Answer 37

:(

Answer 38

Yay

Answer 39

Now I understand the method of computation for these problems.

That helps me understand where I should start. Lol.

Answer 40

Since we have the roots, we can just write it as factored since plugging in alpha or beta will give us 0.\[\Large x^2+\pi x+1=(\alpha - x) (\beta - x)\]

Now all we need to do is plug in -1 to get: \[\Large (-1)^2+\pi (-1)+1=(\alpha +1) (\beta +1)\] So there we have it, -pi+2.

Answer 41

Clever xD

Answer 42

Merry christmas =P

Answer 43

Merry christmas :) have good time Kai !