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Ishaan94
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Show that if \(x^4+ax^3+2x^2+bx+1\) has a real solution, then \(a^2+b^2 ≥ 8\).
 2 years ago
 2 years ago
Ishaan94 Group Title
Show that if \(x^4+ax^3+2x^2+bx+1\) has a real solution, then \(a^2+b^2 ≥ 8\).
 2 years ago
 2 years ago

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Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
@mukushla @KingGeorge
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

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

mukushla Group TitleBest 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
 2 years ago

mukushla Group TitleBest 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: \[164ab \le0\\ab \ge4\\a^2+b^2\ge2ab \ge8\]
 2 years ago

mukushla Group TitleBest 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
 2 years ago

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

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
thanks a lot mukushla
 2 years ago
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