Question

Theory of equations

Answer 1

Let alpha , beta , gamma be the roots of (x-a)(x-b)(x-c) =d , (d not equal to 0)
, then the roots of the equation \[(x-\alpha)(x-\beta)(x-\gamma) +d =0\]
are?

Answer 2

\[x ^{3} + (ab+ac+bc)x -[x ^{2}(a+b+c)] -abc =d \]

Answer 3

@ganeshie8

Answer 4

x^3 - x^2(a+b+c)+x(ab+bc+ca) -abc -d = 0
α+β+γ =-( a+b+c )
αβ+βγ+γβ = ab+bc +ca
αβγ= -(abc+d) or abc = -(d + αβγ)

Answer 5

Since \(\alpha, \beta, \gamma \) are roots of first equation :

\( (x-a)(x-b)(x-c)-d = (x-\alpha)(x-\beta)(x-\gamma)\)

\( \implies (x-a)(x-b)(x-c) = (x-\alpha)(x-\beta)(x-\gamma)+ d\)
\(\implies \) roots of second equation area \(a, b, c\)

Answer 6

the roots of the required equaion is a,b,c

Answer 7

Yes the roots are indeed a,b,c
Thank you both of you !
I got it

Answer 8

yw

Answer 9

I got what what you did @matricked but didn't get hpow you arrvived at the conclusion that a,b,c are the roots

Answer 10

I got using the realtion between the roots and the coefficients..

Answer 11

I mean i got how u did
x^3 - x^2(a+b+c)+x(ab+bc+ca) -abc -d = 0
α+β+γ =-( a+b+c )
αβ+βγ+γβ = ab+bc +ca
αβγ= -(abc+d) or abc = -(d + αβγ)

But didn't how u arrived at the conclusion

Answer 12

first expand (x−α)(x−β)(x−γ)+d =0

( a+b+c ) =-(α+β+γ )
ab+bc +ca = αβ+βγ+γβ
abc = -(d + αβγ)

Answer 13

Ah , i get it thanks