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x-2-2x-9-9-x-1-2-0-please-




Question Number 76214 by Emmanuel_N last updated on 25/Dec/19
x^2 +2x−9+(9/((x+1)^2 ))=0  please
$$\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{9}+\frac{\mathrm{9}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{please} \\ $$
Answered by benjo last updated on 25/Dec/19
(x+1)^2  −10+9/(x+1)^2  =0  let (x+1)^(2 )  =t. now equation t^(2  )  −10t+9=0  this easy to solve
$$\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{10}+\mathrm{9}/\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:=\mathrm{0} \\ $$$$\mathrm{let}\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}\:} \:=\mathrm{t}.\:\mathrm{now}\:\mathrm{equation}\:\mathrm{t}^{\mathrm{2}\:\:} \:−\mathrm{10t}+\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$
Answered by john santuy last updated on 25/Dec/19
x+1=±1.→x=−1±1  x+1=±3. x→−1±3
$${x}+\mathrm{1}=\pm\mathrm{1}.\rightarrow{x}=−\mathrm{1}\pm\mathrm{1} \\ $$$${x}+\mathrm{1}=\pm\mathrm{3}.\:{x}\rightarrow−\mathrm{1}\pm\mathrm{3} \\ $$
Answered by MJS last updated on 25/Dec/19
(x+1)^2 −10+(9/((x+1)^2 ))=0  ⇒ x≠−1  ((x+1)^2 )^2 −10(x+1)^2 +9=0  (x+1)^2 =1∨(x+1)^2 =9  (x+1)^2 =1 ⇒ x+1=±1 ⇒ x=−2∨x=0  (x+1)^2 =9 ⇒ x+1=±3 ⇒ x=−4∨x=2
$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{10}+\frac{\mathrm{9}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:{x}\neq−\mathrm{1} \\ $$$$\left(\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{10}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{9}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}\vee\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}\:\Rightarrow\:{x}+\mathrm{1}=\pm\mathrm{1}\:\Rightarrow\:{x}=−\mathrm{2}\vee{x}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{9}\:\Rightarrow\:{x}+\mathrm{1}=\pm\mathrm{3}\:\Rightarrow\:{x}=−\mathrm{4}\vee{x}=\mathrm{2} \\ $$

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