Question

fun question :)

Answer 1

Let l and m be real numbers such that \[l \neq 0\] . Prove that not all the roots of
\[lx^4 + mx^3 + x^2 + x+1 = 0\] can be real.

Answer 2

fun question huh ? O.o

Answer 3

\[lx^4 + mx^3 + x^2 + x+1 = l(x^2+ax+b)(x^2+cx+d)\]

It is sufficient if we show the discriminant of one of those quadratic factors is less than \(0\)

Answer 4

yes @ganeshie8 :)

Answer 5

this is more better :
\[x^4 + px^3 + qx^2 + qx+q = (x^2+ax+b)(x^2+cx+d)\]

Answer 6

i hav not read about such questions till now...these questions belong to the syllabus of which class? @ganeshie8

Answer 7

It's a real mess!!

Answer 8

class 11th :)

Answer 9

yeah lets try alternatives

Answer 10

what if we take the derivative...would it be helpful?

Answer 11

can we take derivative of 0?

Answer 12

there is a very short method to solve this problem :)

Answer 13

hint-if this equation is hard to work with then try to convert the equation :)

Answer 14

Let \(f(x)=lx^4 + mx^3 + x^2 + x+1 \)


then
\(\begin{align}f(1/x) &= \frac{l}{x^4} + \frac{m}{x^3}+\frac{1}{x^2}+\frac{1}{x}+1\\~\\
&=x^4(l+mx+x^2+x^3+x^4)
\end{align}\)

clearly if the polynomial \(f(1/x)\) has four real roots, then so does the polynomial, \(g(x)=l+mx+x^2+x^3+x^4\), and vice versa.

Next consider the sum of squares of roots of \(g(x)\) :
\(\sum a^2 = \left(\sum a\right)^2- 2\sum ab = (-1)^2-2(1)=-1\lt 0\).

However the sum of squares of real numbers cannot be negative, so it follows that the roots of \(g(x)\) are not all real.

Answer 15

correct @ganeshie8 :)

Answer 16

@ganeshie8 I don't get how \(f(1/x) = x^4(l+mx +x^2+x^3+x^4)\) . Please explain me.

Answer 17

in this step he putted 1/x in the function f(x) :)

Answer 18

sorry, it is a typo, should be :

\[f(1/x) = \frac{1}{x^4}(l+mx +x^2+x^3+x^4)\]

Answer 19

yes, now, it makes sense

Answer 20

thnks for catching :)

Answer 21

ok lol i didn't notice that typo XD :)

Answer 22

@ganeshie8
Suggestion: change your nick to "genius8" :)

Answer 23

^