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Question-37324

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Question Number 37324 by behi83417@gmail.com last updated on 11/Jun/18
Commented by math khazana by abdo last updated on 12/Jun/18
b) ((√(1+x))/(1+(√x)))=x ⇒ (√(1+x)) =x(1+(√x)) with x≥0if x real 1+x =x^2 (1+2(√x) +x) ⇒1+x=x^(3 ) +x^2 +2x^2 (√x) (1+x−(x^3 +x))^2 =4x^5 ⇒ (1+x)^2 −2(1+x)(x^3 +x) +(x^(3 ) +x)^2 −4x^5 =0⇒ x^2 +2x+1 −2(x^3 +x +x^4 +x^2 ) +x^6 +2x^4 +x^2 −4x^5 =0 ⇒ x^6 −4x^5 +2x^4 +2x^2 +2x+1−2x^3 −2x −2x^4 −2x^2 =0 ⇒ x^6 −4x^5 −2x^3 +2x+1 =0 the roots of tbepolynom p(x)=x^6 −4x^5 −2x^3 +2x +1 are x_1 ∼4,1105 (real) x_2 ∼0,8572(real) x_3 ∼i (complex) x_4 ∼−i(complex) x_5 ∼−0,4839 +0,2229i (complex) x_6 ∼−0,4839 −0,2229i(complex) .
$$\left.{b}\right)\:\frac{\sqrt{\mathrm{1}+{x}}}{\mathrm{1}+\sqrt{{x}}}={x}\:\Rightarrow\:\sqrt{\mathrm{1}+{x}}\:={x}\left(\mathrm{1}+\sqrt{{x}}\right)\:{with}\:{x}\geqslant\mathrm{0}{if}\:{x}\:{real} \\ $$$$\mathrm{1}+{x}\:={x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}\sqrt{{x}}\:+{x}\right)\:\Rightarrow\mathrm{1}+{x}={x}^{\mathrm{3}\:} +{x}^{\mathrm{2}} \:+\mathrm{2}{x}^{\mathrm{2}} \sqrt{{x}} \\ $$$$\left(\mathrm{1}+{x}−\left({x}^{\mathrm{3}} +{x}\right)\right)^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{5}} \:\Rightarrow \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+{x}\right)\left({x}^{\mathrm{3}} \:+{x}\right)\:+\left({x}^{\mathrm{3}\:} +{x}\right)^{\mathrm{2}} \:−\mathrm{4}{x}^{\mathrm{5}} =\mathrm{0}\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}\:−\mathrm{2}\left({x}^{\mathrm{3}} \:+{x}\:+{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \right)\:+{x}^{\mathrm{6}} \:+\mathrm{2}{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \\ $$$$−\mathrm{4}{x}^{\mathrm{5}} \:=\mathrm{0}\:\Rightarrow \\ $$$${x}^{\mathrm{6}} \:−\mathrm{4}{x}^{\mathrm{5}} \:+\mathrm{2}{x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} \:−\mathrm{2}{x}\:−\mathrm{2}{x}^{\mathrm{4}} \:−\mathrm{2}{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{6}} \:−\mathrm{4}{x}^{\mathrm{5}} \:−\mathrm{2}{x}^{\mathrm{3}} \:+\mathrm{2}{x}+\mathrm{1}\:=\mathrm{0}\: \\ $$$${the}\:{roots}\:{of}\:{tbepolynom} \\ $$$${p}\left({x}\right)={x}^{\mathrm{6}} \:−\mathrm{4}{x}^{\mathrm{5}} \:−\mathrm{2}{x}^{\mathrm{3}} \:+\mathrm{2}{x}\:+\mathrm{1}\:{are} \\ $$$${x}_{\mathrm{1}} \sim\mathrm{4},\mathrm{1105}\:\left({real}\right) \\ $$$${x}_{\mathrm{2}} \sim\mathrm{0},\mathrm{8572}\left({real}\right) \\ $$$${x}_{\mathrm{3}} \:\sim{i}\:\left({complex}\right) \\ $$$${x}_{\mathrm{4}} \sim−{i}\left({complex}\right) \\ $$$${x}_{\mathrm{5}} \:\sim−\mathrm{0},\mathrm{4839}\:+\mathrm{0},\mathrm{2229}{i}\:\left({complex}\right) \\ $$$${x}_{\mathrm{6}} \:\sim−\mathrm{0},\mathrm{4839}\:−\mathrm{0},\mathrm{2229}{i}\left({complex}\right)\:. \\ $$
Commented by math khazana by abdo last updated on 12/Jun/18
we can also divide p(x) by x^2 +1 and we get a polynom with degre 4 in wich we find the roots...
$${we}\:{can}\:{also}\:{divide}\:{p}\left({x}\right)\:{by}\:{x}^{\mathrm{2}} \:+\mathrm{1}\:{and}\:{we}\:{get} \\ $$$${a}\:{polynom}\:{with}\:{degre}\:\mathrm{4}\:{in}\:{wich}\:{we}\:{find}\:{the}\:{roots}… \\ $$
Commented by behi83417@gmail.com last updated on 12/Jun/18
thank you very much.
$${thank}\:{you}\:{very}\:{much}. \\ $$

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