IIT JEE , AIEEE Forum - Mathematics , Algebra - how 2 find out no. of imaginary roots..plz....

pals_eam

find the no. of least imaginary roots of the eq. x9 + 5x8 - x3 + 7x + 2 = 0

a. 2 b. 4 c. 6 d. 8

the hint was given as ----

total no. of roots - no. of real roots

here the total no. of roots is 9

how do we find the no. of real roots?

plz. help

ashish_banga

is the answer b, that is 4

ziauddin75

answer will be 4(no need of using hint)

dont forget to rte

ziauddin75

We all find no. of real solutions for a 2 degree equation....

but wat if the ask u like a polynomial of degree 8 , they ask u to find the no. of real roots...+ive or -ive.... or complex roots....

dont worry...herez a sol. (some of u might know ) but rest.....look here

for eg. equation is

x4 - 3x + 1 = 0....

we can write it as

(+) x4 - (3x) + (1) = 0...

see the sign change.....

+ to - , then - to + , 2 changes na

so no. of +ive solutions = 2

now to -ive solutions

we gotf(x) = x4 + 3x + 1 = 0....

take f(-x) = x4 + 3x +1 = 0.....

no sign changes......so no -ive root

for complex roots , see the highest degree.....i.e 4 here

so complex roots = highest degree - (no. of + ive + no. of -ive roots)

= 4 - (2+0) = 2 complex roots

hope u got it....ok so now post the no. of +ive , -ive n complex roots =

x8 - 4x7 + 3x6 + x5 - x4 + x3 - 16x + 2 = 0

1 thing more....

if equation of type ax2 + b |x| + c = 0 , no. of real solutions is = 0
if a,b,c>0
and when roots of an equation are equal in magnitude but of opp. sign then sum of roots is = to zero......

given by himanshu

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