341Shouldnt it be 2a_1^2 - 5a_0 a_2 <0
Think its a A . Dasgupta question --- anyhow a good one --- there are two methods actually -- lets see
Prove that the eqn . a0 x5 + a1 x4 + a2 x3 + a3 x2 + a4 x + a5 = 0 cannot have all real roots if 2 a12 - 5 a2 < 0.
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12 Answers
1hari sir , can u give a small hint...as i am not getting any clue to proceed
341If the roots are real then you must have \sum_{i<j} (\alpha_i - \alpha_j)^2 \ge 0
23then p(x) = a0(x-x1)(x-x2)....(x-x5)
whre x1,x2,...x5 are roots
thus
\frac{p(x)}{(x-x_{1})}= (x-x_{2})...(x-x_{5})
now taking limit as x tends to x1
\lim_{x->x_{1}} \frac{p(x)}{(x-x_{1})}= (x-x_{2})...(x-x_{5})
apply L' Hospital to LHS
\lim_{x->x_{1}} \frac{p'(x)}{1}= (x_{1}-x_{2})...(x_{1}-x_{5})
{p'(x_{1})}= (x_{1}-x_{2})...(x_{1}-x_{5})
prophet sir how do we come from here to ur hint ?
by the way i have learnt this from xyz :)
1Extremely sorry , prophet sir , you are absolutely correct , please take a0 = 1 .
1If 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 a12 being less than 5 a2 . 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 x2 + 24 a1 x + 6a2 = 0 ,
1To have all real roots , the discriminant should be >= 0.
But
D = 24 a12 - 360 x 4 a2 = 24 x 12 ( 2 a12 - 5 a2 ) < 0 ,
as given that 2 a12 < 5 a2 .
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 .
341As in my prev post if all the roots are real \sum_{1\le i<j \le 5} (\alpha_i - \alpha_j)^2 \ge 0
\Rightarrow 4 \sum \alpha_i^2 \ge 2 \sum \alpha_i \alpha_j \Rightarrow \sum \alpha_i^2 \ge \frac{1}{2} \sum \alpha_i \alpha_j
Adding 2\sum \alpha_i \alpha_j to both sides we get
\left(\sum \alpha_i)^2 \ge \frac{5}{2} \sum \alpha_i \alpha_j
Now substituting in terms of coefficients, we get2a_1^2 - 5a_0a_2 \ge 0
Hence if 2a_1^2 - 5a_0a_2 <0 we cannot have all roots real
1There was another question related to this , but I couldn't answer it .
The question was to show that if 2a12 < 5a2 , then the eqn. would at most have 2 imaginary roots .
Can anybody answer this please ?