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Ishaan94

  • 2 years ago

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

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  1. Ishaan94
    • 2 years ago
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    @mukushla @KingGeorge

  2. satellite73
    • 2 years ago
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    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

  3. satellite73
    • 2 years ago
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    seems that if either \(a=3\) or \(b=3\) you get a zero at \(x=-1\)

  4. mukushla
    • 2 years ago
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    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

  5. mukushla
    • 2 years ago
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    a better way to proof the second part we want to have \[ax+\frac{b}{x}\le-4\] 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: \[16-4ab \le0\\ab \ge4\\a^2+b^2\ge2ab \ge8\]

  6. mukushla
    • 2 years ago
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    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

  7. Ishaan94
    • 2 years ago
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    Perfect. I was trying to complete square but didn't get it. :/

  8. Ishaan94
    • 2 years ago
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    thanks a lot mukushla

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