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Question-42543

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Question Number 42543 by solihin last updated on 27/Aug/18
Commented by solihin last updated on 27/Aug/18
question 1 please
$${question}\:\:\mathrm{1}\:\:{please} \\ $$
Commented by maxmathsup by imad last updated on 27/Aug/18
1) where this equation is defined ? we must have 2x+1≥0 and 12x^2 +12x +7 = →Δ^′ =6^2 −12×7<0 ⇒∀x 12x^2 +12x+7>0 so we must have x≥−(1/2) (e)⇒ 12 x^2 +12x +7 =(2x+1)^4 =(4x^2 +4x+1)^2 ⇒ 12x^2 +12x +7 =(4x^2 +4x)^2 +2(4x^2 +4x) +1 ⇒ 12x^2 +12x +7 =16(x^4 +2x^3 +x^2 ) +8x^2 +8x +1 ⇒ 16x^4 +32x^3 +24x^2 +8x +1−12x^2 −12x−7 ⇒ 16x^4 +32x^3 +12x^2 −4x−6 =0⇒8x^4 +16x^3 +6x^2 −2x−3 =0 the roots of this equation are x_1 ∼−1,5 (real) x_2 ∼−0,5 +0,5 i(complex) x_3 ∼−0,5−0,5 i (complex) x_4 ∼0,5 (real) x_1 is not solution because x_1 <−(1/2) so the unique slution is x_4 ∼(1/2) .
$$\left.\mathrm{1}\right)\:{where}\:{this}\:{equation}\:{is}\:{defined}\:?\:\:{we}\:{must}\:{have}\:\mathrm{2}{x}+\mathrm{1}\geqslant\mathrm{0}\:{and} \\ $$$$\mathrm{12}{x}^{\mathrm{2}} \:+\mathrm{12}{x}\:+\mathrm{7}\:=\:\:\rightarrow\Delta^{'} =\mathrm{6}^{\mathrm{2}} −\mathrm{12}×\mathrm{7}<\mathrm{0}\:\Rightarrow\forall{x}\:\:\:\mathrm{12}{x}^{\mathrm{2}} \:+\mathrm{12}{x}+\mathrm{7}>\mathrm{0}\:{so}\:{we}\:{must} \\ $$$${have}\:{x}\geqslant−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({e}\right)\Rightarrow\:\mathrm{12}\:{x}^{\mathrm{2}} \:+\mathrm{12}{x}\:\:+\mathrm{7}\:=\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{4}} \:=\left(\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}{x}+\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{12}{x}^{\mathrm{2}} \:+\mathrm{12}{x}\:+\mathrm{7}\:=\left(\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}{x}\right)^{\mathrm{2}} \:+\mathrm{2}\left(\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}{x}\right)\:+\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{12}{x}^{\mathrm{2}} \:+\mathrm{12}{x}\:+\mathrm{7}\:=\mathrm{16}\left({x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \right)\:+\mathrm{8}{x}^{\mathrm{2}} \:+\mathrm{8}{x}\:+\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{16}{x}^{\mathrm{4}} \:+\mathrm{32}{x}^{\mathrm{3}} \:+\mathrm{24}{x}^{\mathrm{2}} \:+\mathrm{8}{x}\:+\mathrm{1}−\mathrm{12}{x}^{\mathrm{2}} \:−\mathrm{12}{x}−\mathrm{7}\:\Rightarrow \\ $$$$\mathrm{16}{x}^{\mathrm{4}} \:+\mathrm{32}{x}^{\mathrm{3}} \:+\mathrm{12}{x}^{\mathrm{2}} \:−\mathrm{4}{x}−\mathrm{6}\:=\mathrm{0}\Rightarrow\mathrm{8}{x}^{\mathrm{4}} \:+\mathrm{16}{x}^{\mathrm{3}} \:+\mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}\:=\mathrm{0} \\ $$$${the}\:{roots}\:{of}\:{this}\:{equation}\:{are}\: \\ $$$${x}_{\mathrm{1}} \sim−\mathrm{1},\mathrm{5}\:\:\left({real}\right) \\ $$$${x}_{\mathrm{2}} \sim−\mathrm{0},\mathrm{5}\:+\mathrm{0},\mathrm{5}\:{i}\left({complex}\right) \\ $$$${x}_{\mathrm{3}} \sim−\mathrm{0},\mathrm{5}−\mathrm{0},\mathrm{5}\:{i}\:\left({complex}\right) \\ $$$${x}_{\mathrm{4}} \:\sim\mathrm{0},\mathrm{5}\:\:\left({real}\right)\:\:\:\:\:\:{x}_{\mathrm{1}} \:{is}\:{not}\:{solution}\:{because}\:{x}_{\mathrm{1}} <−\frac{\mathrm{1}}{\mathrm{2}}\:\:{so}\:{the}\:{unique}\:{slution} \\ $$$${is}\:{x}_{\mathrm{4}} \:\sim\frac{\mathrm{1}}{\mathrm{2}}\:\:. \\ $$
Commented by maxmathsup by imad last updated on 27/Aug/18
after verification we see that x_0 =(1/2) is a exact solution for this equation.
$${after}\:{verification}\:{we}\:{see}\:{that}\:\:{x}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}\:{is}\:{a}\:{exact}\:{solution}\:{for}\:{this}\:{equation}. \\ $$

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