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Question Number 11500 by @ANTARES_VY last updated on 27/Mar/17
(x^2 +x−4)(x^2 +x+4)=9 find the equation has multiple roots.
$$\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}−\mathrm{4}\right)\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{4}\right)=\mathrm{9} \\ $$$$\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{equation}}\:\:\boldsymbol{\mathrm{has}}\:\:\boldsymbol{\mathrm{multiple}}\:\:\boldsymbol{\mathrm{roots}}. \\ $$
Answered by ridwan balatif last updated on 27/Mar/17
((x^2 +x)−4)((x^2 +x)+4)=9 (x^2 +x)^2 −16=9 (x^2 +x)^2 −25=0 ((x^2 +x)−5)((x^2 +x)+5)=0 (i) x^2 +x−5=0 D=b^2 −4ac =1^2 −4(1)(−5) =21 x_(1,2) =((−b±(√D))/(2a)) =((−1±(√(21)))/2) x_1 =((−1+(√(21)))/2) x_2 =−(((1+(√(21)))/2)) (ii) x^2 +x+5=0 D=b^2 −4ac =1^2 −4(1)(5) =−19 (IMPOSSIBLE) point (ii) has no roots
$$\left(\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)−\mathrm{4}\right)\left(\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)+\mathrm{4}\right)=\mathrm{9} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)^{\mathrm{2}} −\mathrm{16}=\mathrm{9} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)^{\mathrm{2}} −\mathrm{25}=\mathrm{0} \\ $$$$\left(\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)−\mathrm{5}\right)\left(\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)+\mathrm{5}\right)=\mathrm{0} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{D}=\mathrm{b}^{\mathrm{2}} −\mathrm{4ac} \\ $$$$\:\:\:\:=\mathrm{1}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{5}\right) \\ $$$$\:\:\:\:=\mathrm{21} \\ $$$$\mathrm{x}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{b}\pm\sqrt{\mathrm{D}}}{\mathrm{2a}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{−\mathrm{1}\pm\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{2}} =−\left(\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{D}=\mathrm{b}^{\mathrm{2}} −\mathrm{4ac} \\ $$$$\:\:\:\:=\mathrm{1}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{5}\right) \\ $$$$\:\:\:\:=−\mathrm{19}\:\left(\mathrm{IMPOSSIBLE}\right) \\ $$$$\mathrm{point}\:\left(\mathrm{ii}\right)\:\mathrm{has}\:\mathrm{no}\:\mathrm{roots} \\ $$
Commented by @ANTARES_VY last updated on 27/Mar/17
The answer is: −5.
$$\boldsymbol{\mathrm{The}}\:\:\boldsymbol{\mathrm{answer}}\:\:\boldsymbol{\mathrm{is}}:\:−\mathrm{5}. \\ $$
Commented by @ANTARES_VY last updated on 27/Mar/17
error
$$\boldsymbol{\mathrm{error}} \\ $$
Commented by mrW1 last updated on 27/Mar/17
please explain what you mean with your question.
$${please}\:{explain}\:{what}\:{you}\:{mean}\:{with} \\ $$$${your}\:{question}. \\ $$
Commented by mrW1 last updated on 27/Mar/17
then your question should be: find the product of the real roots of the equation.
$${then}\:{your}\:{question}\:{should}\:{be}:\:{find}\:{the} \\ $$$${product}\:{of}\:{the}\:{real}\:{roots}\:{of}\:{the} \\ $$$${equation}. \\ $$
Commented by @ANTARES_VY last updated on 27/Mar/17
x_1 ×x_2 =?
$$\boldsymbol{\mathrm{x}}_{\mathrm{1}} ×\boldsymbol{\mathrm{x}}_{\mathrm{2}} =? \\ $$
Commented by mrW1 last updated on 27/Mar/17
yes. x_1 ×x_2 =−5
$${yes}.\:\:{x}_{\mathrm{1}} ×{x}_{\mathrm{2}} =−\mathrm{5} \\ $$
Commented by sandy_suhendra last updated on 27/Mar/17
from the answer of Mr Ridwan Balatif x^2 +x+5=0 has 2 imaginary roots because D<0 x^2 +x−5=0 has 2 real roots x_1 and x_2 because D>0 so x_1 .x_2 = (c/a) = ((−5)/1) = −5
$$\mathrm{from}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{of}\:\mathrm{Mr}\:\mathrm{Ridwan}\:\mathrm{Balatif} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{5}=\mathrm{0}\:\mathrm{has}\:\mathrm{2}\:\mathrm{imaginary}\:\mathrm{roots}\:\mathrm{because}\:\mathrm{D}<\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{5}=\mathrm{0}\:\mathrm{has}\:\mathrm{2}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} \:\mathrm{because}\:\mathrm{D}>\mathrm{0}\:\:\:\:\: \\ $$$$\mathrm{so}\:\mathrm{x}_{\mathrm{1}} .\mathrm{x}_{\mathrm{2}} \:=\:\frac{\mathrm{c}}{\mathrm{a}}\:=\:\frac{−\mathrm{5}}{\mathrm{1}}\:=\:−\mathrm{5} \\ $$
Answered by ajfour last updated on 27/Mar/17
x=((−1±(√(1±20)))/2) .
$$\mathrm{x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}\pm\mathrm{20}}}{\mathrm{2}}\:. \\ $$

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