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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0taxing my brain first thing in the morning one thought: we know generally what this looks like. if it is to have at least one real solution, then at least one zero of the derivative must have second coordinate below the \(x\) axis let me see if i can make that lead to anything

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0seems that if either \(a=3\) or \(b=3\) you get a zero at \(x=1\)

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.3here is my solution but i dont like it i will try to find a better one let \[y=ax+\frac{b}{x}\] u have \[x^4+ax^3+2x^2+bx+1=x^4+2x^2+1+x^2(ax+\frac{b}{x})=(x^2+1)^2+x^2y=0\] then \[y=(x+\frac{1}{x})^2 \] there is at least one solution for equation above if and only if \[y \le 4 \ \ \ \ since \ \ \ x+\frac{1}{x}\ge2\] then we must have \[\max \left\{ y=ax+\frac{b}{x} \right\} =4 \\ y'=0 \\ x=\sqrt{\frac{b}{a}} \\ so \\ \max \left\{ y \right\}=2\sqrt{ab}=4 \\ ab=4 \\ a^2+b^2\ge 2ab=8\] a and be must have the same sign

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.3a better way to proof the second part we want to have \[ax+\frac{b}{x}\le4\] multiply it by x u will get \[ax^2+4x+b \ge0 \\ or \\ ax^2+4x+b \le0 \] since u dont know about the sign of x But In both cases discriminant of quadratic must be less than or equal to zero this gives: \[164ab \le0\\ab \ge4\\a^2+b^2\ge2ab \ge8\]

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.3here is a better one \[x^4 +ax^3+2x^2+bx+1 = (x^2 + \frac{a}{2} x)^2 + (1+\frac{ b}{2} x)^2 + \frac{1}{4} (8−a^2 −b^2)x^2\] In this case, the polynomial is strictly positive unless a^2+b^2−8 ≤ 0

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0Perfect. I was trying to complete square but didn't get it. :/
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