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Question-201081

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Question Number 201081 by Calculusboy last updated on 29/Nov/23
Answered by Rasheed.Sindhi last updated on 29/Nov/23
AnOther Way 2x^3 −x^2 −22x−24=0 let the two roots are 3a & 4a where a≠0 •2(3a)^3 −(3a)^2 −22(3a)−24=0 54a^3 −9a^2 −66a−24=0....(i) •2(4a)^3 −(4a)^2 −22(4a)−24=0 128a^3 −16a^2 −88a−24=0....(ii) (ii)−(i): 74a^3 −7a^2 −22a=0 74a^2 −7a−22=0 [a≠0] (2a+1)(37a−22)=0 a=−(1/2) , (( 22 )/(37)) (false)^★ Two roots are 3(−(1/2)) & 4(−(1/2)) −(3/2) & −2 let the third root is t −(3/2)−2+t=(1/2)^(Sum of the roots) ∧ (−(3/2))(−2)(t)=12^(Product of the roots) ⇒t=4 ^★ two roots are 3(((22)/(37))) & 4(((22)/(37))) ((66)/(37)) & ((88)/(37)) let the third root is t ((66)/(37))+((88)/(37))+t=(1/2) ∧ (((66)/(37)))(((88)/(37)))(t)=12 t has no value(satisfying both) ∴ a=((22)/(37)) is false.
$$\mathrm{AnOther}\:\mathrm{Way} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{22}{x}−\mathrm{24}=\mathrm{0} \\ $$$${let}\:{the}\:{two}\:{roots}\:{are}\:\mathrm{3}{a}\:\&\:\mathrm{4}{a}\:{where}\:{a}\neq\mathrm{0} \\ $$$$\bullet\mathrm{2}\left(\mathrm{3}{a}\right)^{\mathrm{3}} −\left(\mathrm{3}{a}\right)^{\mathrm{2}} −\mathrm{22}\left(\mathrm{3}{a}\right)−\mathrm{24}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{54}{a}^{\mathrm{3}} −\mathrm{9}{a}^{\mathrm{2}} −\mathrm{66}{a}−\mathrm{24}=\mathrm{0}….\left({i}\right) \\ $$$$\bullet\mathrm{2}\left(\mathrm{4}{a}\right)^{\mathrm{3}} −\left(\mathrm{4}{a}\right)^{\mathrm{2}} −\mathrm{22}\left(\mathrm{4}{a}\right)−\mathrm{24}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{128}{a}^{\mathrm{3}} −\mathrm{16}{a}^{\mathrm{2}} −\mathrm{88}{a}−\mathrm{24}=\mathrm{0}….\left({ii}\right) \\ $$$$\left({ii}\right)−\left(\boldsymbol{{i}}\right): \\ $$$$\mathrm{74}{a}^{\mathrm{3}} −\mathrm{7}{a}^{\mathrm{2}} −\mathrm{22}{a}=\mathrm{0} \\ $$$$\mathrm{74}{a}^{\mathrm{2}} −\mathrm{7}{a}−\mathrm{22}=\mathrm{0}\:\:\left[{a}\neq\mathrm{0}\right] \\ $$$$\left(\mathrm{2}{a}+\mathrm{1}\right)\left(\mathrm{37}{a}−\mathrm{22}\right)=\mathrm{0} \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}}\:,\:\frac{\:\mathrm{22}\:}{\mathrm{37}}\:\left({false}\right)^{\bigstar} \\ $$$$\:\:{Two}\:{roots}\:{are}\:\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\&\:\mathrm{4}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{3}}{\mathrm{2}}\:\&\:−\mathrm{2} \\ $$$$\:\:\:\:\:{let}\:{the}\:{third}\:{root}\:{is}\:{t} \\ $$$$\:\:\:\:\:\:\:\:\overset{{Sum}\:{of}\:{the}\:{roots}} {−\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{2}+{t}=\frac{\mathrm{1}}{\mathrm{2}}}\:\wedge\:\overset{{Product}\:{of}\:{the}\:{roots}} {\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)\left(−\mathrm{2}\right)\left({t}\right)=\mathrm{12}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\Rightarrow{t}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:^{\bigstar} \:{two}\:{roots}\:{are}\:\:\mathrm{3}\left(\frac{\mathrm{22}}{\mathrm{37}}\right)\:\&\:\mathrm{4}\left(\frac{\mathrm{22}}{\mathrm{37}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{66}}{\mathrm{37}}\:\&\:\frac{\mathrm{88}}{\mathrm{37}} \\ $$$$\:\:\:\:\:{let}\:{the}\:{third}\:{root}\:{is}\:{t} \\ $$$$\:\:\:\:\frac{\mathrm{66}}{\mathrm{37}}+\frac{\mathrm{88}}{\mathrm{37}}+{t}=\frac{\mathrm{1}}{\mathrm{2}}\:\wedge\:\:\left(\frac{\mathrm{66}}{\mathrm{37}}\right)\left(\frac{\mathrm{88}}{\mathrm{37}}\right)\left({t}\right)=\mathrm{12} \\ $$$$\:\:\:\:\:\:\:{t}\:{has}\:{no}\:{value}\left({satisfying}\:{both}\right) \\ $$$$\:\:\:\:\therefore\:{a}=\frac{\mathrm{22}}{\mathrm{37}}\:{is}\:{false}. \\ $$
Commented by Calculusboy last updated on 29/Nov/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by Sutrisno last updated on 29/Nov/23
x_1 =(3/4)x_2 x_1 +x_2 +x_3 =(1/2) (3/4)x_2 +x_2 +x_3 =(1/2)→x_3 =(1/2)−(7/4)x_2 x_1 .x_2 .x_3 =12 (3/4)x_3 .x_2 ((1/2)−(7/4)x_2 )=12 21x_2 ^3 −6x_2 ^2 +192=0 (x_2 +2)(21x_2 ^2 −48x_2 +96)=0 x_2 =−2 x_1 =−(3/2) x_3 =4
$${x}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{4}}{x}_{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}{x}_{\mathrm{2}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\rightarrow{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{7}}{\mathrm{4}}{x}_{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} .{x}_{\mathrm{2}} .{x}_{\mathrm{3}} =\mathrm{12} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}{x}_{\mathrm{3}} .{x}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{7}}{\mathrm{4}}{x}_{\mathrm{2}} \right)=\mathrm{12} \\ $$$$\mathrm{21}{x}_{\mathrm{2}} ^{\mathrm{3}} −\mathrm{6}{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{192}=\mathrm{0} \\ $$$$\left({x}_{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{21}{x}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{48}{x}_{\mathrm{2}} +\mathrm{96}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{2}} =−\mathrm{2} \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3}} =\mathrm{4} \\ $$
Commented by Calculusboy last updated on 29/Nov/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by mr W last updated on 29/Nov/23
x_1 =3k x_2 =4k x_1 +x_2 +x_3 =(1/2) ⇒x_3 =(1/2)−7k x_1 x_2 +(x_1 +x_2 )x_3 =−((22)/2)=−11 12k^2 +7k((1/2)−7k)=−11 74k^2 −7k−22=0 ⇒k=((7±81)/(148))=−(1/2), ((22)/(37)) x_1 x_2 x_3 =12k^2 ((1/2)−7k)=((24)/2)=12 ⇒k^2 ((1/2)−7k)=1 only k=−(1/2) is ok ⇒x_1 =−(3/2), x_2 =−2, x_3 =4
$${x}_{\mathrm{1}} =\mathrm{3}{k} \\ $$$${x}_{\mathrm{2}} =\mathrm{4}{k} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{7}{k} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} +\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right){x}_{\mathrm{3}} =−\frac{\mathrm{22}}{\mathrm{2}}=−\mathrm{11} \\ $$$$\mathrm{12}{k}^{\mathrm{2}} +\mathrm{7}{k}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{7}{k}\right)=−\mathrm{11} \\ $$$$\mathrm{74}{k}^{\mathrm{2}} −\mathrm{7}{k}−\mathrm{22}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{7}\pm\mathrm{81}}{\mathrm{148}}=−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{22}}{\mathrm{37}} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} =\mathrm{12}{k}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{7}{k}\right)=\frac{\mathrm{24}}{\mathrm{2}}=\mathrm{12} \\ $$$$\Rightarrow{k}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{7}{k}\right)=\mathrm{1} \\ $$$${only}\:{k}=−\frac{\mathrm{1}}{\mathrm{2}}\:{is}\:{ok} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =−\frac{\mathrm{3}}{\mathrm{2}},\:{x}_{\mathrm{2}} =−\mathrm{2},\:{x}_{\mathrm{3}} =\mathrm{4} \\ $$
Commented by Calculusboy last updated on 29/Nov/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Answered by Rasheed.Sindhi last updated on 30/Nov/23
Let the roots are 3k,4k & λk { ((3k+4k+λk=1/2)),((3k.4k + 4k.λk + λk.3k = −11)),((3k.4k.λk=12)) :} ⇒ { ((k(7+λ)=1/2)),((k^2 (12 + 4λ + 3λ) = −11)),((k^3 (12λ)=12⇒λ=(1/k^3 ))) :} ⇒ { ((k(7+(1/k^3 ))=(1/2)⇒14k^3 −k^2 +2=0)),((k^2 (12 + 7((1/k^3 ))) = −11⇒12k^3 +11k+7=0)) :} ⇒ { (((2k+1)(7k^2 −4k+2)=0)),(((2k+1)(6k^2 −3k+7)=0)) :}⇒2k+1=0⇒k=−(1/2) ⇒λ=(1/k^3 )=(1/((−1/2)^3 ))=−8 Roots are: 3(−(1/2)) , 4(−(1/2)) , −8(−(1/2)) −(3/2) , −2 , 4
$${Let}\:{the}\:{roots}\:{are}\:\mathrm{3}{k},\mathrm{4}{k}\:\&\:\lambda{k} \\ $$$$\begin{cases}{\mathrm{3}{k}+\mathrm{4}{k}+\lambda{k}=\mathrm{1}/\mathrm{2}}\\{\mathrm{3}{k}.\mathrm{4}{k}\:+\:\mathrm{4}{k}.\lambda{k}\:+\:\lambda{k}.\mathrm{3}{k}\:=\:−\mathrm{11}}\\{\mathrm{3}{k}.\mathrm{4}{k}.\lambda{k}=\mathrm{12}}\end{cases}\: \\ $$$$\Rightarrow\begin{cases}{{k}\left(\mathrm{7}+\lambda\right)=\mathrm{1}/\mathrm{2}}\\{{k}^{\mathrm{2}} \left(\mathrm{12}\:+\:\mathrm{4}\lambda\:+\:\mathrm{3}\lambda\right)\:=\:−\mathrm{11}}\\{{k}^{\mathrm{3}} \left(\mathrm{12}\lambda\right)=\mathrm{12}\Rightarrow\lambda=\frac{\mathrm{1}}{{k}^{\mathrm{3}} }}\end{cases}\: \\ $$$$\Rightarrow\begin{cases}{{k}\left(\mathrm{7}+\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{14}{k}^{\mathrm{3}} −{k}^{\mathrm{2}} +\mathrm{2}=\mathrm{0}}\\{{k}^{\mathrm{2}} \left(\mathrm{12}\:+\:\mathrm{7}\left(\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\right)\right)\:=\:−\mathrm{11}\Rightarrow\mathrm{12}{k}^{\mathrm{3}} +\mathrm{11}{k}+\mathrm{7}=\mathrm{0}}\end{cases}\:\: \\ $$$$\Rightarrow\begin{cases}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{7}{k}^{\mathrm{2}} −\mathrm{4}{k}+\mathrm{2}\right)=\mathrm{0}}\\{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{6}{k}^{\mathrm{2}} −\mathrm{3}{k}+\mathrm{7}\right)=\mathrm{0}}\end{cases}\Rightarrow\mathrm{2}{k}+\mathrm{1}=\mathrm{0}\Rightarrow{k}=−\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}}{{k}^{\mathrm{3}} }=\frac{\mathrm{1}}{\left(−\mathrm{1}/\mathrm{2}\right)^{\mathrm{3}} }=−\mathrm{8} \\ $$$${Roots}\:{are}:\:\:\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:,\:\:\mathrm{4}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:,\:\:−\mathrm{8}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{3}}{\mathrm{2}}\:,\:−\mathrm{2}\:\:,\:\mathrm{4} \\ $$
Commented by Calculusboy last updated on 01/Dec/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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