Solve this Quartic Equation $x^4+3x^3+4x^2+2x+1=0$. I have tried with various possible solution methods. But I can't find the answer.
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Try subtracting this polynomial from (x+1)^4 – Yunus Syed Oct 13 '15 at 02:01
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All is roots are complex. Even the real and imaginary parts are irrational. So very elementary method will not work. How much do you know on solving quartics ? – Shailesh Oct 13 '15 at 02:01
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See quartic formula. – Lucian Oct 13 '15 at 10:21
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$LHS> x^2(x+\tfrac32)^2+(x+1)^2> 0$, so no real roots... – Macavity Oct 13 '15 at 12:12
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Consider the function and its derivatives $$f(x)=x^4+3 x^3+4 x^2+2 x+1$$ $$f'(x)=4 x^3+9 x^2+8 x+2$$ $$f''(x)=12 x^2+18 x+8$$ The second derivative does not show any real root and then it is always positive (so, at most, two real roots). This implies that the first derivative can only cancel once.
Using Cardano method, the first derivative cancels at $$x_*=\frac{1}{4} \left(-3-\frac{5^{2/3}}{\sqrt[3]{3 \left(9+4 \sqrt{6}\right)}}+\frac{\sqrt[3]{5 \left(9+4 \sqrt{6}\right)}}{3^{2/3}}\right)\approx -0.39417$$ and $$f( -0.39417)\approx 0.673553$$ which is the minimum value of the function.
So, no real root to $f(x)=0$.
Claude Leibovici
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