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imqwerty
 one year ago
fun question :)
imqwerty
 one year ago
fun question :)

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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1Let 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0fun question huh ? O.o

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6\[lx^4 + mx^3 + x^2 + x+1 = l(x^2+ax+b)(x^2+cx+d)\] It is sufficient if we show the discriminant of one of those quadratic factors is less than \(0\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6this is more better : \[x^4 + px^3 + qx^2 + qx+q = (x^2+ax+b)(x^2+cx+d)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i hav not read about such questions till now...these questions belong to the syllabus of which class? @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6yeah lets try alternatives

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what if we take the derivative...would it be helpful?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can we take derivative of 0?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1there is a very short method to solve this problem :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1hintif this equation is hard to work with then try to convert the equation :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Let \(f(x)=lx^4 + mx^3 + x^2 + x+1 \) 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 \(g(x)\) : \(\sum a^2 = \left(\sum a\right)^2 2\sum ab = (1)^22(1)=1\lt 0\). However the sum of squares of real numbers cannot be negative, so it follows that the roots of \(g(x)\) are not all real.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1correct @ganeshie8 :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 I don't get how \(f(1/x) = x^4(l+mx +x^2+x^3+x^4)\) . Please explain me.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1in this step he putted 1/x in the function f(x) :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6sorry, it is a typo, should be : \[f(1/x) = \frac{1}{x^4}(l+mx +x^2+x^3+x^4)\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1yes, now, it makes sense

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6thnks for catching :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1ok lol i didn't notice that typo XD :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 Suggestion: change your nick to "genius8" :)
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