IIT JEE , AIEEE Forum - Mathematics , Algebra

trisha

Show that the equation x⁴+4x³-4x-13=0 has two real roots only.

vpunithreddy

let da roots b a,b,c,d

a+b+c+d=-4

square on both sides and after simplification u get a²+b²+c²+d²=8

a,b,c,d cannot be natural numbers 8 cant b written as sum of 4 squares satisfying da given conditions

den roots cant be fractions i v carefully inspect da given data

so der must b atleast one complex number as root.but as complex roots exist in pairs der shud be 2 complex roots.

so der r 2 real roots

sorry 4 doing a lot of guess work

trisha

vpunithraddy, i think ur proof is not accurate.

can any one please help me?

amar.gupta

Dear,

f(x) = x⁴+4x³-4x-13

f'(x) = 4x³+12x²-4

f''(x) = 12x²+24x

f''(x) = 12x(x+2)

hence f''(x)>0 when x belongs to (-infinity , -2) U ( 0 , infinity)

and f''(x) < 0 when x belongs to (-2,0)

now let f'(x) has three roots a,b,c

a lies between (-infinity , -2 ) as f'(x) changes sign between (-infinity,-2) and f''(a)>0 so minima of f(x)at x=a

b lies between (-2 , 0 ) as f'(x) changes sign between ( -2, 0 )                   and f''(b)<0 so maxima of f(x)at x=b

c lies between (0 , infinity ) as f'(x) changes sign between ( 0 , infinity) and f''(c)>0 so minima of f(x)at x=c

hence f(x) has two minima at x=a,c and one maxima at x=b
and f(x) > 0 when x tends to + infinity and x tends to -infinity

now only three conditions are possible:

a) f(x) has all four real roots.    then f(a)<0 and f(b) >0 and f(c) < 0

b) f(x) has all four imaginary roots. then f(a) >0 and f(b) > 0 and f(c) > 0

c) f(x) has two real and two imaginary roots. then f(a) <0 and f(b) <0 and f(c) < 0

so its sufficient if we proof that f(b)<0

now f(b) =b⁴+4b³-4b-13

we know f'(b) =4b³+12b²-4 =0 or b³+3b²-1=0.......[1]

so f(b) = b(b³+3b²-1) +b³-3b-13       from eq.[1]

or f(b) = b³-(3b+13)                  now as b lies in (-2,0) , (3b+13)>0

hence f(b) < 0

hence there is only two real roots .

trisha

mr.amar gupta in ur solution u have taken that f¹(x)=0 has three real roots.but how can we take like that it may have one real and two imaginary roots.what is the solution then?

trisha

okay sir now i understood how it is.

THANK u sir

trisha

this means r u saying that f¹(x)=0 has three real roots?

but by using cardons method i am not getting real roots for f¹(x).please tell me sir.............

amar.gupta

Dear Trisha,

I have shown you that f'(x) has three real roots as f'(x) has changing its sign in three regions........

(-infinity , -2 ) (-2,0) and (0,infinity)

any continuous function should cut the x axis or in other words it has roots if it changes sign .

f'(x) <0 when x-> -infinity but f'(x) > 0 when x= -2

again f'(x)<0 when x= 0 and f'(x)>0 when x-> infinity

so three must be three real roots.

devesh_l2k007

good njob aman

devesh_l2k007

after uv

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