Question

Let a and b denote the roots of the equation x^2 - 10x + 6 =0 let sn whre n a positive integer stands for a^n + b^n

s(n+2)=? s4=? s4-s2s3=?

Answer 1

if v find the roots of d equation then the question will b a cake walk? so let us focus on finding d roots first

Answer 2

or wait s4= a^4+b^4=(a^2+b^2)^2-2(ab)^2
=((a+b)^2-2ab)^2-2(ab)^2
now sum of roots(a+b)= 10 nd product ab=6, put these values nd u will find ur answer

Answer 3

did nt understand lol

Answer 4

the answer for s4 = 7672......

Answer 5

see this was kind of transformation, so u need to understand it , it is in simplest language, what i did is turned the function into a function of (a+b) nd ab, coz we know these values

Answer 6

so@ @dg123 by solving that i wont get 7672

Answer 7

u know what are the values of a+b and ab?

Answer 8

a^4 + b^4 = (a^2)^2 + (b^2)^2
=(a^2)^2 + (b^2)^2 + 2a^2 b^2 -2a^2 b^2
=(a^2 -b^2)^2 + 2(a^2

Answer 9

@mukushla we can get that from the eq

Answer 10

@shubhamsrg i want to know how this came s4= a^4+b^4=(a^2+b^2)^2-2(ab)^2
=((a+b)^2-2ab)^2-2(ab)^2

Answer 11

aha i got it....

Answer 12

then wat is s(n+2)=?

Answer 13

\[a^{n+2}+b^{n+2}\]

Answer 14

nothing to do with it

Answer 15

ok then s4-s2s3=?

Answer 16

just notice
\[s2=a^2+b^2=(a+b)^2-2ab \\ and \ \ \ s3=a^3+b^3=(a+b)(a^2+b^2-ab)=(a+b)((a+b)^2-3ab)\]

Answer 17

got it thxx