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Show that if \(x^4+ax^3+2x^2+bx+1\) has a real solution, then \(a^2+b^2 ≥ 8\).
 one year ago
 one year ago
Show that if \(x^4+ax^3+2x^2+bx+1\) has a real solution, then \(a^2+b^2 ≥ 8\).
 one year ago
 one year ago

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Ishaan94Best ResponseYou've already chosen the best response.0
@mukushla @KingGeorge
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
taxing 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
seems that if either \(a=3\) or \(b=3\) you get a zero at \(x=1\)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.3
here 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
 one year ago

mukushlaBest ResponseYou've already chosen the best response.3
a 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: \[164ab \le0\\ab \ge4\\a^2+b^2\ge2ab \ge8\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.3
here 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
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Perfect. I was trying to complete square but didn't get it. :/
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
thanks a lot mukushla
 one year ago
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