TargetIIT

Think its a A . Dasgupta question --- anyhow a good one --- there are two methods actually -- lets see

Prove that the eqn . a[ss]0[/ss] x[p]5[/p] + a[ss]1[/ss] x[p]4[/p] + a[ss]2[/ss] x[p]3[/p] + a[ss]3[/ss] x[p]2[/p] + a[ss]4[/ss] x + a[ss]5[/ss] = 0 cannot have all real roots if                  2 a[ss]1[/ss][p]2[/p] - 5 a[ss]2[/ss] < 0.

Shouldnt it be [im]http://codecogs.izyba.com/gif.latex?2a_1%5E2%20-%205a_0%20a_2%20%3C0[/im]

hari sir , can u give a small hint...as i am not getting any clue to proceed

If the roots are real then you must have [im]http://codecogs.izyba.com/gif.latex?%5Csum_%7Bi%3Cj%7D%20%28%5Calpha_i%20-%20%5Calpha_j%29%5E2%20%5Cge%200[/im]

then p(x) = a[ss]0[/ss](x-x[ss]1[/ss])(x-x[ss]2[/ss])....(x-x[ss]5[/ss])

whre x1,x2,...x5 are roots

thus

[im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7Bp%28x%29%7D%7B%28x-x_%7B1%7D%29%7D%3D%20%28x-x_%7B2%7D%29...%28x-x_%7B5%7D%29[/im]

now taking limit as x tends to x[ss]1[/ss]

[im]http://codecogs.izyba.com/gif.latex?%5Clim_%7Bx-%3Ex_%7B1%7D%7D%20%5Cfrac%7Bp%28x%29%7D%7B%28x-x_%7B1%7D%29%7D%3D%20%28x-x_%7B2%7D%29...%28x-x_%7B5%7D%29[/im]

apply L' Hospital to LHS


[im]http://codecogs.izyba.com/gif.latex?%5Clim_%7Bx-%3Ex_%7B1%7D%7D%20%5Cfrac%7Bp%27%28x%29%7D%7B1%7D%3D%20%28x_%7B1%7D-x_%7B2%7D%29...%28x_%7B1%7D-x_%7B5%7D%29[/im]



[im]http://codecogs.izyba.com/gif.latex?%7Bp%27%28x_%7B1%7D%29%7D%3D%20%28x_%7B1%7D-x_%7B2%7D%29...%28x_%7B1%7D-x_%7B5%7D%29[/im]
prophet sir how do we come from here to ur hint ?
[hide]by the way i have learnt this from xyz :)[/hide]

Extremely sorry , prophet sir , you are absolutely correct , please take a[ss]0[/ss] = 1 .

If f ( x ) is a polynomial of degree n having n real roots , and it is differentiable as well as

continuing , then k-th derivative of f ( x ) will have n - k real roots --- [ think why ]

Let’s consider that the given cond. is nothing but a load of rubbish , and the eqn. has all real

roots despite 2 a[ss]1[/ss][p]2[/p] being less than 5 a[ss]2[/ss] . So the third derivative of

the given eqn . will also have all real roots if the first eqn. has all real roots . Third derivative

gives ,

60 x[p]2[/p] + 24 a[ss]1[/ss] x + 6a[ss]2[/ss] = 0 ,

To have all real roots , the discriminant should be >= 0.

But

D = 24 a[ss]1[/ss][p]2[/p] - 360 x 4 a[ss]2[/ss] = 24 x 12 ( 2 a[ss]1[/ss][p]2[/p] - 5 a[ss]2[/ss] ) < 0 ,

as given that 2 a[ss]1[/ss][p]2[/p] < 5 a[ss]2[/ss] .

So the eqn. doesn’t have all real roots as D < 0 .

Hence 3 - rd derivative doesn’t have all real roots .

But we thought that it did , so our assumption is wrong . Hence the given eqn. doesn’t have all real roots .

As in my prev post if all the roots are real [im]http://codecogs.izyba.com/gif.latex?%5Csum_%7B1%5Cle%20i%3Cj%20%5Cle%205%7D%20%28%5Calpha_i%20-%20%5Calpha_j%29%5E2%20%5Cge%200[/im]

[im]http://codecogs.izyba.com/gif.latex?%5CRightarrow%204%20%5Csum%20%5Calpha_i%5E2%20%5Cge%202%20%5Csum%20%5Calpha_i%20%5Calpha_j%20%5CRightarrow%20%5Csum%20%5Calpha_i%5E2%20%5Cge%20%5Cfrac%7B1%7D%7B2%7D%20%5Csum%20%5Calpha_i%20%5Calpha_j[/im]

Adding [im]http://codecogs.izyba.com/gif.latex?2%5Csum%20%5Calpha_i%20%5Calpha_j[/im] to both sides we get

[im]http://codecogs.izyba.com/gif.latex?%5Cleft%28%5Csum%20%5Calpha_i%29%5E2%20%5Cge%20%5Cfrac%7B5%7D%7B2%7D%20%5Csum%20%5Calpha_i%20%5Calpha_j[/im]

Now substituting in terms of coefficients, we get[im]http://codecogs.izyba.com/gif.latex?2a_1%5E2%20-%205a_0a_2%20%5Cge%200[/im]

Hence if [im]http://codecogs.izyba.com/gif.latex?2a_1%5E2%20-%205a_0a_2%20%3C0[/im] we cannot have all roots real

PROPHET SIR
CAN U EXPLAIN HOW U GT THE FIRST EQUATIOJ IN #4

sum of squares ≥0  needs to be explained?

There was another question related to this , but I couldn't answer it .

The question was to show that if 2a[ss]1[/ss][p]2[/p] < 5a[ss]2[/ss] , then the eqn. would at most have 2 imaginary roots .

Can anybody answer this please ?

As a counterexample:

[im]http://codecogs.izyba.com/gif.latex?P%28x%29%20%5Cequiv%20x%28x%5E2+1%29%28x%5E2+4%29%20%3D%20x%5E5+5x%5E3+4x[/im]