Menu Close

Question-39218




Question Number 39218 by behi83417@gmail.com last updated on 03/Jul/18
Answered by MJS last updated on 12/Aug/18
x_1 =1; y_1 =(√2) (plain to see but there are 3                                more real solutions)  y=−x^2 +1+(√2)  (−x^2 +1+(√2))^2 +x−3=0  x^4 −2(1+(√2))x^2 +x+2(√2)=0  (x−1)(x^3 +x^2 −(1+2(√2))x−2(√2))=0  x_1 =1  x^3 +x^2 −(1+2(√2))x−2(√2)=0  solved by using trigonometric method  (Cardano doesn′t work in case of 3 real  solutions)  x_2 ≈−2.16492; y_2 ≈−2.27265  x_3 ≈−.700407; y_3 ≈1.92364  x_4 ≈1.86532; y_4 ≈−1.06521
$${x}_{\mathrm{1}} =\mathrm{1};\:{y}_{\mathrm{1}} =\sqrt{\mathrm{2}}\:\left(\mathrm{plain}\:\mathrm{to}\:\mathrm{see}\:\mathrm{but}\:\mathrm{there}\:\mathrm{are}\:\mathrm{3}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{more}\:\mathrm{real}\:\mathrm{solutions}\right) \\ $$$${y}=−{x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$\left(−{x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +{x}−\mathrm{3}=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}^{\mathrm{2}} +{x}+\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right){x}−\mathrm{2}\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{1} \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right){x}−\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{solved}\:\mathrm{by}\:\mathrm{using}\:\mathrm{trigonometric}\:\mathrm{method} \\ $$$$\left(\mathrm{Cardano}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{work}\:\mathrm{in}\:\mathrm{case}\:\mathrm{of}\:\mathrm{3}\:\mathrm{real}\right. \\ $$$$\left.\mathrm{solutions}\right) \\ $$$${x}_{\mathrm{2}} \approx−\mathrm{2}.\mathrm{16492};\:{y}_{\mathrm{2}} \approx−\mathrm{2}.\mathrm{27265} \\ $$$${x}_{\mathrm{3}} \approx−.\mathrm{700407};\:{y}_{\mathrm{3}} \approx\mathrm{1}.\mathrm{92364} \\ $$$${x}_{\mathrm{4}} \approx\mathrm{1}.\mathrm{86532};\:{y}_{\mathrm{4}} \approx−\mathrm{1}.\mathrm{06521} \\ $$
Commented by behi83417@gmail.com last updated on 04/Jul/18
dear MJS!thanks in advance.
$${dear}\:{MJS}!{thanks}\:{in}\:{advance}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *