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solve-5-x-2-4x-2-3-x-2-4x-1-4-x-2-4x-8-




Question Number 172027 by Mikenice last updated on 23/Jun/22
solve  (5/(x^2 +4x+2))=(3/(x^2 +4x+1))−(4/(x^2 +4x+8))
$${solve} \\ $$$$\frac{\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}=\frac{\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}} \\ $$
Commented by Mikenice last updated on 23/Jun/22
thanks sir
$${thanks}\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Jun/22
Let x^2 +4x+1=y  (5/(y+1))=(3/y)−(4/(y+7))  5y(y+7)=3(y+1)(y+7)−4y(y+1)  5y^2 +35y=3y^2 +24y+21−4y^2 −4y  6y^2 +15y−21=0  2y^2 +5y−7=0  (y−1)(2y+7)=0  y=1 ∣ y=−(7/2)  x^2 +4x+1=1 ∣ x^2 +4x+1=−(7/2)  x(x+4)=0 ∣ 2x^2 +8x+9=0  (x=0 ∣ x=−4) ∣ x=((−8±(√(64−72)))/4)  (x=0 ∣ x=−4) ∣ x=((−4±(√2))/2)
$${Let}\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}={y} \\ $$$$\frac{\mathrm{5}}{{y}+\mathrm{1}}=\frac{\mathrm{3}}{{y}}−\frac{\mathrm{4}}{{y}+\mathrm{7}} \\ $$$$\mathrm{5}{y}\left({y}+\mathrm{7}\right)=\mathrm{3}\left({y}+\mathrm{1}\right)\left({y}+\mathrm{7}\right)−\mathrm{4}{y}\left({y}+\mathrm{1}\right) \\ $$$$\mathrm{5}{y}^{\mathrm{2}} +\mathrm{35}{y}=\mathrm{3}{y}^{\mathrm{2}} +\mathrm{24}{y}+\mathrm{21}−\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y} \\ $$$$\mathrm{6}{y}^{\mathrm{2}} +\mathrm{15}{y}−\mathrm{21}=\mathrm{0} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} +\mathrm{5}{y}−\mathrm{7}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left(\mathrm{2}{y}+\mathrm{7}\right)=\mathrm{0} \\ $$$${y}=\mathrm{1}\:\mid\:{y}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=\mathrm{1}\:\mid\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${x}\left({x}+\mathrm{4}\right)=\mathrm{0}\:\mid\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{9}=\mathrm{0} \\ $$$$\left({x}=\mathrm{0}\:\mid\:{x}=−\mathrm{4}\right)\:\mid\:{x}=\frac{−\mathrm{8}\pm\sqrt{\mathrm{64}−\mathrm{72}}}{\mathrm{4}} \\ $$$$\left({x}=\mathrm{0}\:\mid\:{x}=−\mathrm{4}\right)\:\mid\:{x}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Commented by Mikenice last updated on 23/Jun/22
thanks sir
$${thanks}\:{sir} \\ $$

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