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Question-76340

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Question Number 76340 by Master last updated on 26/Dec/19
Commented by behi83417@gmail.com last updated on 26/Dec/19
x=−1,0
$$\mathrm{x}=−\mathrm{1},\mathrm{0} \\ $$
Answered by behi83417@gmail.com last updated on 26/Dec/19
−3x^2 +2x+5=(1+x)(−3x+5) ⇒(1+x)(−3x+5)=(√((1+x)(1−x)))(4x^2 +8x+5) 1)[1+x=0⇒x=−1] 2) (1+x)(−3x+5)^2 =(1−x)(4x^2 +8x+5)^2 (1+x)(9x^2 −30x+25)=(1−x)(16x^4 +64x^2 +25+64x^3 +40x^2 +80x) ⇒9x^2 −30x+25+9x^3 −30x^2 +25x= 16x^4 +64x^2 +25+64x^3 +40x^2 +25x −16x^5 −64x^3 −25x−64x^4 −40x^3 −80x^2 ⇒−16x^5 −48x^4 −49x^3 +45x^2 +5x=0 ⇒−x(16x^4 +48x^3 +49x^2 −45x−5)=0 ⇒[x=0] 16x^4 +48x^3 +49x^2 −45x−5=0 ⇒x=−0.1 , 0.623
$$−\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}=\left(\mathrm{1}+\mathrm{x}\right)\left(−\mathrm{3x}+\mathrm{5}\right) \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{x}\right)\left(−\mathrm{3x}+\mathrm{5}\right)=\sqrt{\left(\mathrm{1}+\mathrm{x}\right)\left(\mathrm{1}−\mathrm{x}\right)}\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{5}\right) \\ $$$$\left.\mathrm{1}\right)\left[\mathrm{1}+\boldsymbol{\mathrm{x}}=\mathrm{0}\Rightarrow\boldsymbol{\mathrm{x}}=−\mathrm{1}\right] \\ $$$$\left.\mathrm{2}\right)\:\left(\mathrm{1}+\mathrm{x}\right)\left(−\mathrm{3x}+\mathrm{5}\right)^{\mathrm{2}} =\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{4x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+\mathrm{x}\right)\left(\mathrm{9x}^{\mathrm{2}} −\mathrm{30x}+\mathrm{25}\right)=\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{16x}^{\mathrm{4}} +\mathrm{64x}^{\mathrm{2}} +\mathrm{25}+\mathrm{64x}^{\mathrm{3}} +\mathrm{40x}^{\mathrm{2}} +\mathrm{80x}\right) \\ $$$$\Rightarrow\mathrm{9x}^{\mathrm{2}} −\mathrm{30x}+\mathrm{25}+\mathrm{9x}^{\mathrm{3}} −\mathrm{30x}^{\mathrm{2}} +\mathrm{25x}= \\ $$$$\:\:\:\:\:\:\mathrm{16x}^{\mathrm{4}} +\mathrm{64x}^{\mathrm{2}} +\mathrm{25}+\mathrm{64x}^{\mathrm{3}} +\mathrm{40x}^{\mathrm{2}} +\mathrm{25x} \\ $$$$\:\:\:\:−\mathrm{16x}^{\mathrm{5}} −\mathrm{64x}^{\mathrm{3}} −\mathrm{25x}−\mathrm{64x}^{\mathrm{4}} −\mathrm{40x}^{\mathrm{3}} −\mathrm{80x}^{\mathrm{2}} \\ $$$$\Rightarrow−\mathrm{16x}^{\mathrm{5}} −\mathrm{48x}^{\mathrm{4}} −\mathrm{49x}^{\mathrm{3}} +\mathrm{45x}^{\mathrm{2}} +\mathrm{5x}=\mathrm{0} \\ $$$$\Rightarrow−\mathrm{x}\left(\mathrm{16x}^{\mathrm{4}} +\mathrm{48x}^{\mathrm{3}} +\mathrm{49x}^{\mathrm{2}} −\mathrm{45x}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow\left[\boldsymbol{\mathrm{x}}=\mathrm{0}\right] \\ $$$$\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{48}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{49}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{45}\boldsymbol{\mathrm{x}}−\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{\mathrm{x}}=−\mathrm{0}.\mathrm{1}\:\:,\:\:\mathrm{0}.\mathrm{623} \\ $$
Answered by mr W last updated on 26/Dec/19
−1≤x≤1 (5−3x)(1+x)=(√((1−x)(1+x)))(4(x+1)^2 +1) x=−1 is a solution. for x≠−1 (5−3x)(√(1+x))=(√((1−x)))(4(x+1)^2 +1) (5−3x)^2 (1+x)=(1−x)(4(x+1)^2 +1)^2 let t=x+1 (8−3t)^2 t=(2−t)(4t^2 +1)^2 (64−48t+9t^2 )t=(2−t)(16t^4 +8t^2 +1) 16t^5 −32t^4 +17t^3 −64t^2 +65t−2=0 (t−1)(t^2 +t+2)(16t^2 −32t+1)=0 t=1 ⇒x=0 t=((16±(√(16^2 −16)))/(16))=1±((√(15))/4) ⇒x=±((√(15))/4) t=((−1±i(√7))/2) ⇒x=((−3±i(√7))/2) summary of all solutions: x=0 x=−1 x=±((√(15))/4)≈±0.9682 x=((−3±i(√7))/2)
$$−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\left(\mathrm{5}−\mathrm{3}{x}\right)\left(\mathrm{1}+{x}\right)=\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}\left(\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${x}=−\mathrm{1}\:{is}\:{a}\:{solution}. \\ $$$${for}\:{x}\neq−\mathrm{1} \\ $$$$\left(\mathrm{5}−\mathrm{3}{x}\right)\sqrt{\mathrm{1}+{x}}=\sqrt{\left(\mathrm{1}−{x}\right)}\left(\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left(\mathrm{5}−\mathrm{3}{x}\right)^{\mathrm{2}} \left(\mathrm{1}+{x}\right)=\left(\mathrm{1}−{x}\right)\left(\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$${let}\:{t}={x}+\mathrm{1} \\ $$$$\left(\mathrm{8}−\mathrm{3}{t}\right)^{\mathrm{2}} {t}=\left(\mathrm{2}−{t}\right)\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{64}−\mathrm{48}{t}+\mathrm{9}{t}^{\mathrm{2}} \right){t}=\left(\mathrm{2}−{t}\right)\left(\mathrm{16}{t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{16}{t}^{\mathrm{5}} −\mathrm{32}{t}^{\mathrm{4}} +\mathrm{17}{t}^{\mathrm{3}} −\mathrm{64}{t}^{\mathrm{2}} +\mathrm{65}{t}−\mathrm{2}=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{2}\right)\left(\mathrm{16}{t}^{\mathrm{2}} −\mathrm{32}{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=\mathrm{1}\:\Rightarrow{x}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{16}\pm\sqrt{\mathrm{16}^{\mathrm{2}} −\mathrm{16}}}{\mathrm{16}}=\mathrm{1}\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\:\Rightarrow{x}=\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$$${t}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\Rightarrow{x}=\frac{−\mathrm{3}\pm{i}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$ \\ $$$${summary}\:{of}\:{all}\:{solutions}: \\ $$$${x}=\mathrm{0} \\ $$$${x}=−\mathrm{1} \\ $$$${x}=\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\approx\pm\mathrm{0}.\mathrm{9682} \\ $$$${x}=\frac{−\mathrm{3}\pm{i}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$
Commented by MJS last updated on 26/Dec/19
the complex solutions don′t fit the given equation
$$\mathrm{the}\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{don}'\mathrm{t}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$
Commented by mr W last updated on 26/Dec/19
how to see this? due to (√(1−x^2 )) ?
$${how}\:{to}\:{see}\:{this}? \\ $$$${due}\:{to}\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:? \\ $$
Commented by MJS last updated on 26/Dec/19
yes.
$$\mathrm{yes}. \\ $$
Commented by mr W last updated on 26/Dec/19
thanks sir!
$${thanks}\:{sir}! \\ $$
Answered by MJS last updated on 26/Dec/19
squaring and transforming ⇒ x(x+1)(x^2 +3x+4)(x^2 −((15)/(16)))=0 ⇒ x_1 =−1 x_2 =−((√(15))/4) x_3 =0 x_4 =((√(15))/4) x_(5, 6) =−(3/2)±((√7)/2)i (don′t fit the given equation) other path let x=((2t)/(t^2 +1)) ⇒ 5+((4t)/(t^2 +1))−((12t^2 )/((t^2 +1)^2 ))=((1−t^2 )/(t^2 +1))(((16t^2 )/((t^2 +1)^2 ))+((16t)/(t^2 +1))+5) ⇒ t(t+1)(t^2 +t+2)(t^2 −(3/5))=0 ⇒ t_1 =−1 ⇒ x_1 =−1 t_2 =−((√(15))/5) ⇒ x_2 =−((√(15))/4) t_3 =0 ⇒ x_3 =0 t_4 =((√(15))/5) ⇒ x_4 =((√(15))/4) t_(5, 6) =−(1/2)±((√7)/2)i ⇒ x_(5, 6) =−(3/2)±((√7)/2)i (false)
$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming} \\ $$$$\Rightarrow \\ $$$${x}\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{15}}{\mathrm{16}}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} =−\mathrm{1} \\ $$$${x}_{\mathrm{2}} =−\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$$${x}_{\mathrm{3}} =\mathrm{0} \\ $$$${x}_{\mathrm{4}} =\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$$${x}_{\mathrm{5},\:\mathrm{6}} =−\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\mathrm{i}\:\left(\mathrm{don}'\mathrm{t}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\right) \\ $$$$ \\ $$$$\mathrm{other}\:\mathrm{path} \\ $$$$\mathrm{let}\:{x}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow \\ $$$$\mathrm{5}+\frac{\mathrm{4}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{12}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}\left(\frac{\mathrm{16}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{16}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\mathrm{5}\right) \\ $$$$\Rightarrow \\ $$$${t}\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{2}\right)\left({t}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{5}}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${t}_{\mathrm{1}} =−\mathrm{1}\:\Rightarrow\:{x}_{\mathrm{1}} =−\mathrm{1} \\ $$$${t}_{\mathrm{2}} =−\frac{\sqrt{\mathrm{15}}}{\mathrm{5}}\:\Rightarrow\:{x}_{\mathrm{2}} =−\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$$${t}_{\mathrm{3}} =\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{3}} =\mathrm{0} \\ $$$${t}_{\mathrm{4}} =\frac{\sqrt{\mathrm{15}}}{\mathrm{5}}\:\Rightarrow\:{x}_{\mathrm{4}} =\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$$${t}_{\mathrm{5},\:\mathrm{6}} =−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\mathrm{i}\:\Rightarrow\:{x}_{\mathrm{5},\:\mathrm{6}} =−\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\mathrm{i}\:\left(\mathrm{false}\right) \\ $$

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