IIT JEE , AIEEE Forum - Mathematics , Algebra

vahiniprasad

(x-a)^3 +(x-b)^3+(x-c)^3=0

how many real & imaginary roots?

iitjee08aspirant

wat r the comditions on a,b and c???????????

b_srikalyan009

we dont need any condition

Nithy

i think u have to expand it, arrange in increasing order of coefficients. let it be

f(x) no of changes in sign as u move from left to right gives no of positive roots.

substitite -x ie f(-x). no of change in sign: no of -ve real roots

3- sum of above. is no of imaginary roots.

for ur question

3x³ - x²(3a - 3b - 3c) + x 3( a² + b² + c ²) - (a³ + b ³+ c³)=0

( constant do not matter. don't contribute to sign)

no of + roots: 3

- ve : 0

im= 3-3 =0.

ans: 3 real roots no imaginary roots.

rate me if i am i correct????

b_srikalyan009

if u differentiate the exuation u get a positive quantity i.e 3*((x-a)²+ (x-b)²+ (x-c)²))

.hence the function is increasing.

the range of the function is from neg.infinity to +ve infinity.

hence it can touch y=0 at only one point.

we dont require any condition on a,b,c

metal

Expand the expression.

you will get 3x³-3x²(a+b+c)+3x(a²+b²+c²)-(a³+b³+c³)

It is clear from this equation that there are 3 roots, none of which are imaginary.

please rate me........

rajatsen91

3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)
This is a third degree equation having all real roots

b_srikalyan009

rajatsen see my method where did i go wrong?

it should definately have 1 real root and 2 imaginary roots

vahiniprasad

thank u srikalyan009
yep this is the right ans

metal

You were right srikalyan.................

Good job.....................

b_srikalyan009

so please rate me