imqwerty
  • imqwerty
fun question :)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
imqwerty
  • 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.
sohailiftikhar
  • sohailiftikhar
fun question huh ? O.o
ganeshie8
  • 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 \(0\)

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imqwerty
  • imqwerty
yes @ganeshie8 :)
ganeshie8
  • ganeshie8
this is more better : \[x^4 + px^3 + qx^2 + qx+q = (x^2+ax+b)(x^2+cx+d)\]
anonymous
  • anonymous
i hav not read about such questions till now...these questions belong to the syllabus of which class? @ganeshie8
Loser66
  • Loser66
It's a real mess!!
imqwerty
  • imqwerty
class 11th :)
ganeshie8
  • ganeshie8
yeah lets try alternatives
anonymous
  • anonymous
what if we take the derivative...would it be helpful?
anonymous
  • anonymous
can we take derivative of 0?
imqwerty
  • imqwerty
there is a very short method to solve this problem :)
imqwerty
  • imqwerty
hint-if this equation is hard to work with then try to convert the equation :)
ganeshie8
  • 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.
imqwerty
  • imqwerty
correct @ganeshie8 :)
Loser66
  • Loser66
@ganeshie8 I don't get how \(f(1/x) = x^4(l+mx +x^2+x^3+x^4)\) . Please explain me.
imqwerty
  • imqwerty
in this step he putted 1/x in the function f(x) :)
ganeshie8
  • ganeshie8
sorry, it is a typo, should be : \[f(1/x) = \frac{1}{x^4}(l+mx +x^2+x^3+x^4)\]
Loser66
  • Loser66
yes, now, it makes sense
ganeshie8
  • ganeshie8
thnks for catching :)
imqwerty
  • imqwerty
ok lol i didn't notice that typo XD :)
Loser66
  • Loser66
@ganeshie8 Suggestion: change your nick to "genius8" :)
imqwerty
  • imqwerty
^

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