## imqwerty one year ago fun question :)

1. imqwerty

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.

2. sohailiftikhar

fun question huh ? O.o

3. ganeshie8

$lx^4 + mx^3 + x^2 + x+1 = l(x^2+ax+b)(x^2+cx+d)$ It is sufficient if we show the discriminant of one of those quadratic factors is less than $$0$$

4. imqwerty

yes @ganeshie8 :)

5. ganeshie8

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

6. anonymous

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

7. Loser66

It's a real mess!!

8. imqwerty

class 11th :)

9. ganeshie8

yeah lets try alternatives

10. anonymous

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

11. anonymous

can we take derivative of 0?

12. imqwerty

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

13. imqwerty

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

14. ganeshie8

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.

15. imqwerty

correct @ganeshie8 :)

16. Loser66

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

17. imqwerty

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

18. ganeshie8

sorry, it is a typo, should be : $f(1/x) = \frac{1}{x^4}(l+mx +x^2+x^3+x^4)$

19. Loser66

yes, now, it makes sense

20. ganeshie8

thnks for catching :)

21. imqwerty

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

22. Loser66

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

23. imqwerty

^