Matematické Fórum

stuart clark

Consider the equation in terms of $y$ as $y^4-2y^3+2y^2-y+q=0$ for real number $q$

The sum of non real roots can be

$(a)\; 1$

$(b)\; 1/2$

$(c)\; 1/4$

$(d)$ None.

also evaluation of minimum number of non real roots of the equation is

kerajs · 02. 02. 2020 07:25

$W(y)=y^4-2y^3+2y^2-y+q=(y-y_1)(y-y_2)(y-y_3)(y-y_4) \Rightarrow y_1+ y_2+ y_3+ y_4=2 \\ \\ W'(y)=4y^3-6y^2+4y-1=(y-\frac12)(4(y-\frac12)+1) \\ W_{min}=W(\frac12)=\frac{-3}{16}+q\\ \\ a) \ \ q>\frac{3}{16} \ \Rightarrow \ \text{sum of non real roots }=2\\ \\ b) \ \ q \le \frac{3}{16} \\ (y(y-1))^2+(y(y-1))+q=0 \\ y^2-y=\frac{-1-\sqrt{1-4q}}{2} \ \ y^2-y=\frac{-1+\sqrt{1-4q}}{2} \\ b.1) \ \ y^2-y=\frac{-1-\sqrt{1-4q}}{2}\\ \Delta \ge 0 \ \Rightarrow \ \text{sum of real roots }=y_1+y_2=1\\ b.2) \ \ y^2-y=\frac{-1+\sqrt{1-4q}}{2}\\ \Delta < 0 \ \Rightarrow \ \text{sum of non real roots }=y_3+y_4=1$

misaH · 02. 02. 2020 08:12

$W(y)=y^4-2y^3+2y^2-y+q=(y-y_1)(y-y_2)(y-y_3)(y-y_4) \Rightarrow y_1+ y_2+ y_3+ y_4=2 \\ W'(y)=4y^3-6y^2+4y-1=(y-\frac12)(4(y-\frac12)+1) \\ W_{\min}=W$\frac12$=\frac{-3}{16}+q\\ a) \ \ q>\frac{3}{16} \\ \Rightarrow \\ \text{sum of non real roots }=2 \\ b) \ \ q \le \frac{3}{16} \\ (y(y-1))^2+(y(y-1))+q=0 \\ y^2-y=\frac{-1-\sqrt{1-4q}}{2}\\ y^2-y=\frac{-1+\sqrt{1-4q}}{2} \\ b.1) \\y^2-y=\frac{-1-\sqrt{1-4q}}{2}\\ \Delta \ge 0 \\ \Rightarrow \\ \text {sum of real roots }=y_1+y_2=1\\ b.2) \\y^2-y=\frac{-1+\sqrt{1-4q}}{2}\\ \Delta \le 0 \\ \Rightarrow \\ \text{sum of non real roots }=y_3+y_4=1$

stuart clark · 03. 02. 2020 10:37

Thanks ↑ kerajs:↑ misaH:

misaH · 04. 02. 2020 06:27

↑ stuart clark:

:-)

Matematické Fórum

#1 01. 02. 2020 16:37 — Editoval stuart clark (01. 02. 2020 16:37)

polynomials

#2 02. 02. 2020 07:25

Re: polynomials

#3 02. 02. 2020 08:12

Re: polynomials

#4 03. 02. 2020 10:37

Re: polynomials

#5 04. 02. 2020 06:27

Re: polynomials

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