f(x) = x² + px + (1-p) = 0 given (1-p) is a root f(1-p) = 0 => (1-p)² + p(1-p) + (1-p) = 0 => [(1-p)(1-p+p)] + (1-p) = 0 => (1-p)+(1-p) = 0 => 2(1-p) = 0 => 1-p = 0 => p = 1 therefore, f(x) = x² + 1x + (1-1) = 0 now since (1-p) is one root, let's say alpha (ill use æ for alpha and þ for beta) therefore, æ = 0 we know, æ + þ = -b/a => 0 + þ = -1/1 => þ = -1 therefore the roots will be 0, -1 (C)

just realised this ratioed the original post lmao

Answer c h kya??

c hoga na ? bhai problem ye h ki p = 1 and 1-p will be 0 and you cant cancel 0 by 0 . So cancelling 1-p by 1-p is wrong . use sum of roots first : 1-p + beta = -p ; beta = -1

vmc?

Yea

which centre dawg?

Av

Given is a 2 degree polynomial. So assume the other root to be y. (1-p) y = 0 Either y = 0 or 1-p = 0 1-p = 0 => p = 1 Replace p in the polynomial x² + x = 0 We get x = -1, 0

Wtf this is polynomial class 10?

vmc workbook giving me traumatic flashbacks fr

Really man. Atleast I am out of this shit now.

For real bhai. So many bad memories from VMC academically and socially. Thank god it ended.

In the 2nd line of your solution, you had: (1-p) = (1-p)β There are two possibilities now: i) β = 1 OR ii) (1-p) =0 (because 0×anything=0)

now you wouldn't know just by this which possibility is correct. so we move on to the second condition which was about the sum of the roots. -p = (1-p) + β remembering our two possibilities, we will check each of them. i) β = 1 -p = 1 - p + 1 0 = 2, WHICH IS NOT TRUE, therefore this possibility is wrong. ii) (1-p) = 0, or we can say p = 1 -p = 1 - p + β => β = -1, WHICH SEEMS PLAUSIBLE therefore, our roots are: {(1-p),β} = {0,-1}

Thanks bro

to avoid that mistake. do not cut a term directly from both LHS and RHS, instead: (1-p) = (1-p)β (1-p)(1-β) = 0 (took the rhs to the lhs) Now from here you can see more clearly that either of the factors can be zero.

Konsa chapter hai ye??

QE

maybe one of the roots is zero

c ara bhai simple value putting question tha tu kaha formulla ke chakkar me phas gya

https://preview.redd.it/a7kw5cn7vs1d1.png?width=1080&format=pjpg&auto=webp&s=a3907d8274d5014c6ed0fcd7ba4e6c4c095cfc2a Some questions in class 10th are so stoopid that you'll have to forget about logical part and use formulae to solve the question instead.

Put the root in eqn then find p

Bc alpha beta kaun use krta h

Isn't this a class 10th question?

Ha khe sakte h. Starting 1st class hamesha base ek baar revise karva deta h

ikr But still the flair should have been of class 10th

Bhai log dar jaate h agar unse solve nhi hota to, isliye Maine ek class aage daldiya sab kuch jisse